MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3

$\displaystyle \int \cos x\log(\cos x) dx=\sin x \log(\cos x) + \log|\sec x+\tan x|+f(x)+c$ , then

$f(x)=$

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$-\sin x$
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$\cos x$
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$\tan x$
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$\cot x$

$\displaystyle \int f(x)dx=g(x)$ then

$\displaystyle \int x^{3}f(x^{2})dx=$

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$\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})-\int g(x^{2})dx^{2}\}$
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$\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})-\int g(x^{2})dx\}$
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$\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})+\int g(x^{2})dx^{2}\}$
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$\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})+\int g(x^{2})dx\}$

Explanation

$\displaystyle \int x^{3}f(x^{2})dx$

$\displaystyle \frac{1}{2}\int x^{2}f(x^{2})dx^{2}$

$\displaystyle x^{2}=k$

$\displaystyle \frac{1}{2}\int kf(k)dk$

$\displaystyle \frac{1}{2}[k\int f(k)dk-\int (\int f(k)dk)dk]$

$\displaystyle =\frac{1}{2}[kg(k)-\int g(k)dk]$

$\displaystyle \int f(x)dx=\int f(k)dk=g$

$\displaystyle =\frac{1}{2}[x^{2}g(x^{2})-\int g(x^{2})dx^{2}]$

$\displaystyle \int\{f(x)g''(x)-f''(x)g(x)\}dx=$

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$f(x)g^{'} (x)-g(x)f^{'}(x) + c$
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$g(x)f^{'}(x) - f(x)g^{'}(x) + c$
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$g(x)f^{'}(x)+f(x)g^{'}(x) + c$
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$f(x) g(x) + c$

Explanation

$\displaystyle \int [f\ cx)g^{"}(x)-f^{"}(x)g(x)]dx$

$\displaystyle \int f(x)g^{''}(x)+f^{'}(x)g^{'}(x)dx$

$\displaystyle -\int [f^{'}(x)\ g^{'}\ (x)+f^{''}(x)g(x)]dx$

$\displaystyle \int d[f\ cx)g^{'}(x)]-\int d(f^{'}(x)g(x)]$

$\displaystyle =\ f(x)g^{'}(x)-f^{'}(x)\ g(x)+c$

$\displaystyle \int [f(x)g^{''}(x)-f^{''}(x)g(x)]dx$

$\displaystyle =f(x)g^{'}(x)-f^{'}(x)g(x)+c$

$\displaystyle \int (e^\sqrt[3]{x}dx)=$

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$x^{2/3}-2 x^{1/3} +2+c$
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$(x^{2/3} - 2x^{1/3} +2)\exp(\sqrt[3]{x})+c$
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$3(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c$
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$2(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c$

Explanation

$I= \int { { e }^{ \sqrt [ 3 ]{ x } }dx }$

Put

$\sqrt [ 3 ]{ x } =t$

$\displaystyle \dfrac { 1 }{ 3 } { x }^{ -2/3 }dx=dt$

$\displaystyle dx=\dfrac { 3 }{ { t }^{ -2 } } dt$

So,

$I=3\int { { { t }^{ 2 }e }^{ t }dt }$

$=3[{ { t }^{ 2 }e }^{ t }-2\int { t{ e }^{ t }dt } ]+c$

$=3{ { t }^{ 2 }e }^{ t }-6[t{ e }^{ t }-\int { { e }^{ t }dt } ]+c$

$=3{ { t }^{ 2 }e }^{ t }-6t{ e }^{ t }+6{ e }^{ t }+c$

$=3(x^{2/3} - 2x^{1/3}+2)e^{\sqrt[3]{x}}+c$

$\displaystyle \int\sqrt{x}.\log xdx=$

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$\displaystyle \frac{2}{3}x^{3/2}.\log x-\frac{4}{9}x^{3/2}+c$
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$\displaystyle \frac{2}{3}x^{3/2}.\log x+x^{3/2}+c$
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$x^{3/2}.(\displaystyle \log x-\frac{2}{3})+c$
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$\displaystyle \frac{2}{5}x^{3/2}(\log x+1)+c$

Explanation

$\int \sqrt{x}\ logx\ dx$

$=log\ x\ \int \sqrt{x}\ dx$

$-\int 1/x\ \left ( \int \sqrt{x}\ dx \right )dx$

$=2/3\ logx\ x\ ^{3/2}-2/3\int \dfrac{x^{3/2}}{x}dx$

$2/3\ logx(x\ ^{3/2})-2/3\int x\ ^{1/2}\ dx$

$2/3\ logx(x\ ^{3/2})-4/9\ x\ ^{3/2}+c$

$2/3\ x^{3/2}\ log\ x\ 4/9\ x^{3/2}+c$

$\int \sqrt{x}\ logx\ dx=2/3\ x^{3/2}logx-4/9\ x^{3/2}+c$

$f(x)dx =g(x) +c$ , then

$\displaystyle \int f^{-1}(x)dx$ is equal to

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$xf^{-1}(x) +c$
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$f[g^{-1}(x)]+c$
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$x f^{-1}(x)-g[f^{-1}(x)]+c$
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$g^{-1}(x) +c$

Explanation

$\int f(x)dx=g(x)+c$

$\Rightarrow \frac{dg(x)}{dx}=f(x)$

$\int f^{-1}(x)dx$ let

$x=f(y)$

$\Rightarrow \int ydf(y)$

$=y\int df(y)-\int 1[\int df(y)]dy$

$=yf(y)-\int f(y)dy$

$=yf(y)-g(y)+c$

$xf^{-1}(x)-g(f^{-1}(x))+c$
So,

$\int f^{-1}(x)dx=xf^{-1}(x)-g(f^{-1}(x))+c$

Evaluate

$\displaystyle \int e^{\sin x}\sin 2xdx$

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$2e^{\sin x}(\sin x+1)+c$
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$e^{\sin x}(\sin x+2)+c$
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$e^{\sin x}(2\sin x-2)+c$
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$e^{\sin x}(3\sin x -2)+c$

Explanation

$\int e ^{\sin x} \cdot 2 \sin x \cos x dx$

$=2\int e ^{\sin x} \cdot \sin x d(\sin x)$

Take

$\sin x=t$

$=2\int e^{t} \cdot t dt$

We know,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

Now by integrating the functions by parts, by taking

$t$ as

$u$ and

$e^t$ as

$v$ , we get

$=2(t e ^{t}-e ^{t})$

$=2\ e\ ^{t}(t-1)$

Put the value of

$t$

$=2e^{\sin x}(\sin x-1)=e^{\sin x}(2\sin x-2)$

$\Rightarrow \int e\ ^{\sin x}\sin \ 2x\ dx=e^{\sin x}(2\sin x-2)+c$

$\displaystyle \int x\tan x\sec^{2}xdx=$

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$\displaystyle \frac{1}{2}[x\tan^{2} x-tanx +x]+c$
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$\frac{1}{2}[x\tan^{2} x-tanx -x]+c$
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$\displaystyle \frac{1}{2}[x\tan^{2}x+ tanx-x]+c$
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$x\tan^{2} x-tanx+x+c$

Explanation

$L=\int (x\ tanx)sec^{2}x\ dx$

$=x\ tanx\int sec^{2}x\ dx$

$-\int \left ( [tanx+x\ sec^{-2}x]\int sec^{2}x\ dx \right )dx$

$=x\ tanx\ tanx-\int [tan^{2}x+(x\ sec^{2}x\ tan\ x)]dx$

$L=x\ tan^{2}x-\int tan^{2}x\ dx-\int x\ s\ e\ c^{2}x\ tanx\ dx$

$2L=x\ tan^{2}x-\int (sec^{2}x-1)dx$

$2L=x\ tan^{2}x-\ tanx+x+c$

$L=\dfrac{1}{2}\ [x\ tan^{2}x-tanx+x]+c$

$\int x\ tanx\ sec^{2}x\ dx=\dfrac{1}{2}\ [x\ tan^{2}x-tanx+x]+c$

$\displaystyle \int\frac{2x+sin2x}{1+cos2x}dx=$

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$xsin2x+c$
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$-xtanx+c$
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$xsecx+c$
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$xtanx+c$

Explanation

$\int \dfrac{2x+sin2x}{1+cos2x}dx$

$\int \dfrac{2x+sin2x}{2cos^{2}x}dx$

$\int xsec^{2}x^{dx}+\int tanxdx$

$x\int sec^{2}xdx-\int (\int sec^{2}xdx)dx$

$+\int tanxdx$

$x\int sec^{2}xdx-\int tanxdx+\int tanxdx$

$xtanx+c$

$\int \dfrac{2x+sin2x}{1+cos2x}=xtanx+c$

$\displaystyle \int e^{x}(3x^{2}+7x+2)dx=$

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$e^{x}[3x^{2}+x+1] +c$
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$e^{x}[3x^{2}+7x+1] +c$
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$e^{x}[3x^{2} - x- l] +c$
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$e^{x}[3x^{2}+x]+c$

Explanation

$\int e^{x}(3x^{2}+7x+2)dx$

$3\int e^{x}x^{2}\ dx+7\ \int\ e^{x}x\ dx+2\int e^{x}dx$

$\int e^{x}\ x^{2}\ dx=x^{2}e^{x}-2\int xe^{x}\ dx+c$

$\int xe^{x}=xe^{x}-e^{x}+c$

$\int e^{x}=e^{x}+c$

$3[x^{2}\ e^{x}]-6\int xe^{x}\ dx+7\int e^{x}-x\ dx+2e^{x}$

$3\ x^{2}e^{x}+xe^{x}-e^{x}+e^{x}+2e^{x}+c$

$=e^{n}[3x^{2}+x+1]+c$

$\int e^{x}(3x^{2}+7x+2)dx=\ e^{x}[3x^{2}+x+1]+c$

Evaluate

$\displaystyle \int e^{x}(\log x+\frac{1}{x^{2}})dx$

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$e^{x}$ logx $+c$
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$\displaystyle e^{x}(\log x-\frac{1}{x})+c$
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$\displaystyle e^{x}(\log x+\frac{1}{x})+c$
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$\displaystyle \frac{e^{x}}{x^{2}}+c$

Explanation

$\displaystyle \int e^{x}(log x+\dfrac{1}{x^{2}})dx$

$\displaystyle \int e^{x}log x dx+\int e^{x}\dfrac{1}{x^{2}}dx$

$\displaystyle =log x\int e^{x}dx-\int \dfrac{1}{x}e^{x}dx+\int \dfrac{e^{x}}{x^{2}}dx$

$\displaystyle =log x e^{x}-\int \dfrac{1}{x}e^{x}dx+e^{x}\int \dfrac{1}{x^{2}}dx -\int \left(e^{x}\int \dfrac{1}{x^{2}}dx \right) dx$

$\displaystyle =log x e^{x}-\int \dfrac{1}{x}e^{x}dx+e^{x}[-\dfrac{1}{x}]+\int e^{x}\dfrac{1}{x}dx$

$\displaystyle =e^{x}[log x-\dfrac{1}{x}]+c$

So,

$\displaystyle \int e^{x}(log x+\dfrac{1}{x^{2}})dx = e^{x}[log x-\dfrac{1}{x}]+c$

$\displaystyle \int x^{3} \cos x^{2}dx=$

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$x^{2} \sin x^{2} -\cos x^{2}+c$
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$\displaystyle \frac{1}{2} [ x^{2} \sin x^{2}+ \cos x^{2}] +c$
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$\displaystyle \frac{1}{3} [ x^{2} \sin x^{2}+ \cos x^{2}] +c$
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$x^{2} \sin x^{2}+ \cos x^{2}+c$

Explanation

$\displaystyle \int x^{3} \cos x^{2}dx$

Put

$x^{2}=t$

$xdx=\displaystyle \frac{dt}{2}$

$=\displaystyle \frac { 1 }{ 2 } \int t\cos tdt$

$\displaystyle =\frac { 1 }{ 2 } [t\sin { t } -\int { \sin { t } dt } ]$

$\displaystyle =\frac { 1 }{ 2 } t\sin { t } +\frac { 1 }{ 2 } \cos { t } +c$

$=\displaystyle \frac{1}{2} [ x^{2} \sin x^{2}+ \cos x^{2}] +c$

$\displaystyle \int\frac{\log(x^{2}+a^{2})}{x^{2}}dx =$

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$\displaystyle \frac{-\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c$
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$\displaystyle \frac{\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c$
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$\displaystyle \frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-1}(\frac{x}{a})+c$
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$\displaystyle-\frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-l}(\frac{x}{a})+c$

Explanation

$\int \dfrac{log(x^{2}+a^{2})}{x^{2}}dx$

$log(x^{2}+a^{2})\int 1/x^{2}dx-\int (\dfrac{2x}{x^{2}+a^{2}}-[\int \dfrac{1}{x^{2}}dx])dx$

$=log(x^{2}+a^{2})(-1/x)+2\int \dfrac{1}{(x^{2}+a^{2})}dx+c$

$=log (x^{2}+a^{2})(-1/x)+\dfrac{2}{a}tan^{-1}(x/a)+c$

$=\dfrac{-log(x^{2}+a^{2})}{x}+\dfrac{2}{a}tan^{-1}(x/a)+c$

$\int \dfrac{log(x^{2}+a^{2})}{x^{2}}dx=\dfrac{-log(x^{2}+a^{2})}{x}+\dfrac{2}{a}tan^{-1}(x/a)+c$

$\displaystyle \int\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}dx$ is equal to

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$x-\sqrt{1-x^{2}}sin^{-1}x+c$
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$x+\sqrt{1-x^{2}}sin^{-1}x+c$
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$x+sin^{-1}x+c$
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$x-sin^{-1}x+c$

Explanation

Let

$I=\displaystyle \int \dfrac { x\sin ^{ -1 } x }{ \sqrt { 1-x^{ 2 } } } dx$

Put

$\sin ^{ -1 } x=t$

$\implies x=\sin t$

$\dfrac {1}{\sqrt { 1-x^{ 2 } }}dx=dt$

$I=\int t\sin { t } dt$

$=-t\cos { t } +\int { \cos { t } dt }$ ....integration by parts

$=-t\cos { t } +\sin { t } +C$

$=-\sqrt { 1-{ x }^{ 2 } }\sin ^{ -1 } x +x+C$

$=x-\sqrt { 1-{ x }^{ 2 } }\sin ^{ -1 } x +C$

Hence, option 'A' is correct.

$\displaystyle \int e^{sin^{-1}x}[1+\frac{x}{\sqrt{1-x^{2}}}]$ dx

$=$

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$xe^{sin^{-1}x}+c$
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$e^{sin^{-1}x}+c$
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$\displaystyle \frac{1}{\sqrt{1-x^{2}}}e^{\sin^{-1}x}+c$
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$x^{2}e^{sin^{1}X}+c$

Explanation

$\displaystyle I= \int e^{sin^{-1}x}[1+\frac{x}{\sqrt{1-x^{2}}}]dx$

Put

$\sin^{-1} x=t$

$\Rightarrow x=\sin t$

$\Rightarrow dx=\cos t dt$

So,

$\displaystyle I=\int e^{t}(\sin t +\cos t)dt$

$\Rightarrow I = (\sin t +\cos t)e^{t}-\int (\cos t - \sin t)e^{t} dt$

$\Rightarrow I= (\sin t +\cos t)e^{t} - (\cos t-\sin t )e^{t} -\int e^{t}(\sin t +\cos t)dt$

$\Rightarrow 2I= 2\sin t e^{t}+C$

$\Rightarrow I =xe^{\sin^{-1} x}+C$

$\displaystyle \int e^\sqrt{x}dx=$

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$2 e^\sqrt{x}(\sqrt{x}+1)+c$
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$2 e^ \sqrt{x} \left( \sqrt{x}-1 \right) + c$
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$e^ x (x - 1) + c$
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$e ^x (x + 1) + c$

Explanation

$I=\int e^{ \sqrt { x } }dx$
Put

$\sqrt { x } =t$

$\displaystyle \frac { 1 }{ 2 } { x }^{ -1/2 }dx=dt$

$\Rightarrow dx=2tdt$

So,

$I=2\int te^{t}dt$

$=2[t{ e }^{ t }-\int { { e }^{ t }dt } ]$

$= 2t{ e }^{ t }-2{ e }^{ t }+C$

$=2 e^{\sqrt{x} } \left( \sqrt{x}-1 \right)+c$

$\displaystyle \int tan^{-1}\sqrt{\frac{1-x}{1+x}}d{x}=$

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$\displaystyle \frac{1}{2}[x cos^{-1} x-\sqrt{1-x^{2}}]+c$
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$[xcos^{-1}x-\sqrt{1-x^{2}}]+c$
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$\displaystyle \frac{1}{2}[x cos^{-1} x+\sqrt{1-x^{2}}]+c$
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$x cos^{-1} x+\sqrt{1-x^{2}}+c$

Explanation

$\displaystyle I =\int tan^{-1}\sqrt{\frac{1-x}{1+x}}d{x}=$

Put

$x=\cos {\theta}$

$\Rightarrow dx=-\sin{\theta}d{\theta}$

$\displaystyle I =\int tan^{-1}(\tan {\frac{\theta}{2}}) (-\sin{\theta})d{\theta}$

$\Rightarrow \displaystyle I=-\int \frac{\theta}{2} \sin{\theta} d{\theta}$

$\Rightarrow \displaystyle I=-\frac{1}{2}\int \theta\sin {\theta}d{\theta}$

$\Rightarrow \displaystyle I=-\frac{1}{2}[{\theta}(-\cos {\theta})+\sin{\theta}]+c$

$\Rightarrow \displaystyle I=\frac{1}{2}[{\theta}(\cos {\theta})-\sin{\theta}]+c$

$\displaystyle I= \frac{1}{2}[x cos^{-1} x-\sqrt{1-x^{2}}]+c$

(A) :

$\displaystyle \int(2x$ tan x

$\sec^{2}x+\tan^{2}x)dx=x\tan^{2}x+c$
(B) :

$\displaystyle \int[xf^{'}(x)+f(x)]dx=xf(x)+c$

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Both A and R are true and R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true R is false
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A is false but R is true.

$\displaystyle \int\frac{x-\sin x}{1-\cos x}dx=$

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$log |1-Cosx | +c$
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$log | x - sin x | +c$
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$x\displaystyle \tan\frac{x}{2}+c$
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$-x\displaystyle \cot\frac{x}{2}+c$

Explanation

$\int (\dfrac{x-sin x}{1-cos x})dx=\int \dfrac{x-sin x}{2sin^{2}x/2}dx$

$=\int (\dfrac{x}{2})co sec^{2}x/2^{dx}-\int cot x/2 dx$

$=x/2\int cosec^{2}x/2 d(x/2-\int (\int cosec^{2}x/2 dx/2))dx$

$-\int cot x/2 dx$

$=-x cot x/2+c+\int cot x/2 dx-\int cot x/2 dx$

$=-x cot x/2+c$

$\int (\dfrac{x-sin x}{1-cos x})dx=-x cot x/2+c$

$\displaystyle \int(\log x)^{2}dx=$

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$x[(\log x)^{2}-2$ $\log x$ $+2]+c$
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$x[(\log x)^{2}+2$ $\log x$ $+2]+c$
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$[(\log x)^{2}-2$ $\log x$ $+2]+c$
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$[(\log x)^{2}+2$ $\log x$ $+2]+c$

Explanation

$\displaystyle \int (\log x)_{1}^{2}dx \int (\log x)^{2}-1 dx$

$\displaystyle (\log x)^{2} \int 1dx- \int \dfrac{2(\log x)}{x}x dx$

$\displaystyle =(\log x)^{2}[x]-2\int \log x dx$

$\displaystyle \int \log x=x \log x-x$

$\displaystyle =x(\log x)^{2}-2[x \log x-x]+c$

$\displaystyle =x(\log x)^{2}-2x(\log x)+2x+c$

$\displaystyle =x[(\log x)^{2}-2(\log x)+2]+c$

$\displaystyle \Rightarrow \int (\log x)^{2}dx=x[(\log x)^{2}-2(\log x)+2]+c$

$\displaystyle \int x^{n}.\log xdx=$

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$\displaystyle \frac{x^{n+1}}{(n+1)^{2}}(\log x-\frac{1}{(n+1)^{2}})+c$
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$(\displaystyle \log x-\frac{1}{(n+1)^{2}})+c$
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$\displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)})+c$
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$\displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)^{2}})+c$

Explanation

$\int x^{n}log x dx$

$=log x\int x^{n} dx-\int [\dfrac{1}{x}\int x^{n}.dx]dx$

$=log x\dfrac{x^{n+1}}{n+1}-\dfrac{1}{n+1}\int x^{n}.dx$

$=log x\dfrac{x^{n+1}}{n+1}-\dfrac{1}{n+1}[\dfrac{x^{n+1}}{(n+1)}]+c$

$=\dfrac{x^{n+1}}{(n+1)}[logx-\dfrac{1}{(n+1)}]+c$

$\int x^{n} log x dx=\dfrac{x^{n+1}}{n+1}[logx-\dfrac{1}{(n+1)}]+c$

$\displaystyle \int\frac{\log(1-x)}{x^{2}}dx$ equal to

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$[(1+\displaystyle \frac{1}{x})\log(1-x)-\log x]+c$
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$(1-\displaystyle \frac{1}{x})\log(1-x)-\log x+c$
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$(\displaystyle \frac{1}{x}-1)\log(1-x)+\log x+c$
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$(\displaystyle \frac{1}{x}-1)\log(1-x)-\log x+c$

Explanation

$\dfrac{\int log(1-x)}{x^{2}}dx$

$log(1-x)\int 1/x^{2}dx-\int [\dfrac{1}{(1-x)}\int \dfrac{1}{x^{2}}dx ]dx$

$\dfrac{-log(1-x)}{x}+c-\int \dfrac{1}{(1-x)}-\dfrac{1}{x}dx$

$=\dfrac{-log(1-x)}{x}-\int [\dfrac{1}{x}+\dfrac{1}{1-x}]dx+c$

$=\dfrac{-log(1-x)}{x}-log x+log(1-x)+c$

$=log(1-x)[1-\dfrac{1}{x}]-log x+c$

$\int \dfrac{log(1-x)}{x^{2}}dx- log(1-x)[1-1/x]-log x+c$

$\int \sin \sqrt {x}dx =$

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$2(Sin\sqrt {x} - \sqrt {x}.\cos \sqrt {x}) + c$
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$2(Sin \sqrt {x} + \sqrt {x}.\cos \sqrt {x}) + c$
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$2 (Sin \sqrt{x} -\dfrac {1}{2} \sqrt {x} .\cos \sqrt {x}) + c$
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$2(Sin \sqrt {x} + \dfrac {1}{2} \sqrt {x} . \cos \sqrt {x}) + c$

$\displaystyle \int 32x^{3}(\log x)^{2}dx$ is equal to

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$8x^{4}(\log x)^{2}+c$
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$x^{4}\{8(\log x)^{2}-4\log x+1\}+c$
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$x^{4}\{8(\log x)^{2}-4\log x\}+c$
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$x^{3}\{(\log x)^{2}+2\log x\}+c$

Explanation

$\int 32\ x^{3}(log\ x)^{2}= 32 \int x^{3}(log\ x)^{2}dx$

$= 32\ x^{3}\int (log\ x)$

$\left [ (log\ x)^{2} \right ]x^{3}dx - \int \left ( 2 \dfrac{log\ x}{x}\ \dfrac{x^{4}}{4} \right )dx$

$= (log x)^{2} \dfrac{-x^{4}}{4} - \int \ ^{1}/_{2}log\ x . x^{3}-dx$

$= (log\ x)^{2}-\dfrac{x^{4}}{4}-\ ^{1}/_{2} \int log\ x-x^{3}-dx$

$\int log\ x.x^{3} dx= log\ x.\dfrac{x^{4}}{4}-\int \ ^{1}/_{x} .\dfrac{x^{4^{3}}}{4}.dx$

$= log\ x.\dfrac{x^{4}}{4}- \ ^{1}/_{16}\ x^{4}+c$

$32 \left [ (log\ x)^{2}-\dfrac{x^{4}}{4}-log\ x.\dfrac{x^{4}}{8}+\dfrac{1}{32}x^{4} \right ]+c$

$= x^{4}\left [ 8(log\ x)^{2}-4log\ x+1 \right ]+c$

$\displaystyle \int x^{3}(\log x)^{2}dx=$

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$\displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}+\frac{1}{8}]+c$
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$\displaystyle \frac{x^{4}}{4}[(\log x)^{2}+\frac{\log x}{2}+\frac{1}{8}]+c$
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$(\log x)^{2}+2 logx +8+c$
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$\displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}-\frac{1}{8}]+c$

Explanation

$\int x^{3}(log\ x)^{2}dx=(log\ x)^{2}-\int x^{3}dx-\int \left ( 2\dfrac{log\ x}{x}.\dfrac{x^{4}}{4} \right )dx$

$= (log\ x)^{2}-\dfrac{x^{4}}{4}-\int \dfrac{1}{2} (log\ x)(x^{3})dx$

$\int (log\ x)(x^{3})dx= log\ x.\dfrac{x^{4}}{4}-\dfrac{x^{4}}{16}+c$

$\dfrac{x^{4}}{4}\left [ (log\ x)^{2}-\dfrac{log\ x}{2}+\ \dfrac{1}{8} \right ]+c$

$\int \displaystyle \frac{sin\frac{x}{2}+cos^{2}\frac{x}{2}}{1+cos x}dx=$

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$\sec\frac{x}{2}+x+c$
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$\displaystyle \sec\frac{x}{2}+\frac{x}{2}+c$
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$\displaystyle\sec\frac{x}{2}-\frac{x}{2}+C$
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$\displaystyle\sec\frac{x}{2}-X+C$

Explanation

$\int \dfrac{sin\frac{x}{2} +cos^{2}\frac{x}{2}}{1+cos\ x}dx$

$\int \dfrac{sin\frac{x}{2} +cos^{2}\frac {x}{2}}{2cos^{2}\frac{x}{2}}dx$

$= \frac{x}{2}\left [ \int tan \frac{x}{2} sec \frac {x}{2}\frac{dx}{2}+\int 1. dx \right ]$

$sec \frac{x}{2} + x + c$

Solve

$\displaystyle \int e^{\log x}.\cos xdx$

0%
$x \sin x -cos x + c$
0%
$\displaystyle \frac{x}{2}\sin x+\cos x+c$
0%
$x \sin x + cos x + c$
0%
$x \sin x + cos2 x + c$

Explanation

We know,

$e^{log x}=x$

$\int e ^{log x}cos x dx= \int x cos x dx$

We know,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

Now by integrating the functions by parts, taking

$x$ as

$u$ and

$cos x$ as

$v$ , we get

$\int x cos x dx = x \int cos x dx - \int 1(\int cos x)dx$

$=x sin x +cos x +c$

$\int e ^{log x}cos x dx= x \cdot sin x +cos x+c \cdot$

$\displaystyle \int\cos x.\log(\cos x)dx=$

0%
$sin x log (cos x) - log (cos x) + c$
0%
$sin x log (cos x) + sec x + c$
0%
$sin x log (cos x) - sin x + log |sec x + tan x| + c$
0%
$sinx log (cosx) - sec x + c$

Explanation

$\int cos\ x\ log(cos\ x)dx.$

$= log(cos\ x)\int cos\ x\ dx-\int \left [ \dfrac{log(cos\ x)}{dx}-\int (cos\ x)dx \right ]$

$= log(cos\ x)sin\ x-\int \dfrac{(-sin\ x).(t\ sin\ x)}{cos\ x}dx$

$= log(cos\ x)sin\ x+\int \dfrac{sin^{2}x}{cos\ x}\ dx$

$= log(cosx)sinx+\int secx\ dx-\int cosx\ dx$

$= sin\ x\ log(cos\ x)-sin\ x+log\left | sec x+tan\ x \right |+c$

$\displaystyle \int x^{3} sin x^{2}dx=$

0%
$\displaystyle \frac{1}{2}\sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c$
0%
$\displaystyle \frac{1}{2}\sin x^{2}+\frac{1}{2}x^{2}\cos x^{2}+c$
0%
$\sin x^{2}-x^{2}\cos x^{2}+c$
0%
$\displaystyle \sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c$

Explanation

$\int x^{3}sin x^{2}dx$

$=1/2\int x^{2} sin x^{2}dx^{2}, let x^{2}=t$

$=1/2\int t sint dt$

$\int t sint dt=-tcost+sint$

$=1/2\int tsin dt=1/2[sin t-t cost]+c$

$=1/2[sin x^{2}-x^{2}cos x^{2}]+c$

$=1/2sin x^{2}-x^{2}/2 cos x^{2}+c$

$\int x^{3}sin x^{2} dx=1/2sin x^{2}-\dfrac{x^{2}}{2}cos x^{2}+c$

Evaluate

$\displaystyle \int x.\sec^{2}x \ dx$

0%
$x \tan x + \log |\sec x| + c$
0%
$x \tan x - \log |\sec x| + c$
0%
$\displaystyle \frac{x}{2}\tan x-\log |\sec x|+c$
0%
$\dfrac{x}{2}\tan x+\log|\sec x|+c$

Explanation

We know,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

Now by integrating the functions by parts, taking

$x$ as

$u$ and

$sec^2 x$ as

$v$ .

$\displaystyle \int x. \sec ^{2}xdx =x\int \sec ^{2}xdx-\int 1\left(\int \sec ^{2}xdx\right)dx$

$\displaystyle =x \tan x-\int \tan x \ dx+c$

$=x \tan x-\log |\sec x|+c$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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