CBSE Questions for Class 12 Commerce Maths Integrals Quiz 1 - MCQExams.com

We have,

$$I=\int{\dfrac{dx}{\sqrt{9-25{{x}^{2}}}}}$$

$$ I=\int{\dfrac{dx}{5\sqrt{\dfrac{9}{25}-{{x}^{2}}}}} $$

$$ I=\dfrac{1}{5}\int{\dfrac{dx}{\sqrt{{{\left( \dfrac{3}{5} \right)}^{2}}-{{x}^{2}}}}} $$

We know that

$$\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$$

Therefore,

$$ I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{x}{\dfrac{3}{5}} \right)+C $$

$$ I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C $$

Hence, this is the correct answer.

$$\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln  x}}=\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln  x}}$$

$$\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln  x}}=2\sqrt{1+lnx}\displaystyle \displaystyle \vert_{1}^{e^3}$$

$$=2\sqrt{1+3}-2\sqrt{1}$$

$$=2\sqrt{4}-2=2(1)$$

$$=2$$

So,  $$\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln  x}}=2$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos   x }{3+4  \sin x}dx=\int_{0}^{\dfrac{\pi}{3}}\dfrac{d(\sin  x)}{3+4  \sin x}$$

$$\displaystyle =\int_{0}^{\dfrac{\pi}{3}}\dfrac{d (\sin  x) }{3+4  \sin x}=\dfrac{1}{4}\left[\log  (3+4  \sin  x)\right]_{0}^{\dfrac{\pi}{3}}$$

$$\displaystyle =\dfrac{1}{4}\left[\log  \left[3+4\left(\sin \dfrac{\pi}{3}\right)\right]-\log  (3+4  \sin (0))\right]$$

$$=\dfrac{1}{4}\left (\log\left( 3+\dfrac{4\sqrt{3}}{2}\right)-\log  (3)\right)$$

$$=\dfrac{1}{4}(\log  (3+2\sqrt{3})-\log 3)$$

$$=\dfrac{1}{4}\log \left ( \dfrac{3+2\sqrt{3}}{3} \right )$$

$$\therefore \displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos   x}{3+4 \sin  x}dx=\dfrac{1}{4} \log  \left(\dfrac{3+2\sqrt{3}}{3}\right)$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{cos  x}{1+sin^2  x}dx=\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{d  sin  x}{1+sin^2  x}$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{d  sin x}{1+sin^2 x}=\left [( tan^{-1}(sin   x)+ c\right ) \displaystyle ]_{0}^{\dfrac{\pi}{2}}\ (\int\dfrac{1}{1+x^2}=\tan^{-1}x)$$

$$=\left ( tan^{-1}(sin  ( \dfrac{\pi}{2})+ c \right )-\left ( tan^{-1}(sin   0)+ c\right )$$

$$=tan^{-1}(1)$$

$$=\dfrac{\pi}{4}$$

We need to find the value of $$\displaystyle \int_{-1}^{3} \left[ \tan^{-1}\left ( \dfrac{x}{x^2+1} \right) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right] dx$$

We know $$\tan^{-1}p+\cot^{-1}p=\dfrac{\pi}{2}$$

$$\Rightarrow   \tan^{-1}p+\tan^{-1}\dfrac{1}{p}=\dfrac{\pi}{2}$$

So, $$\tan^{-1}\left ( \dfrac{x}{x^2+1} \right )+\tan^{-1}\left ( \dfrac{x^2+1}{x} \right )=\dfrac{\pi}{2}$$

$$\displaystyle \int_{-1}^{3}\left [ \tan^{-1} \left ( \dfrac{x}{1+x^2}\right ) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right ]dx$$

$$=\displaystyle \int_{-1}^{3}\dfrac{\pi}{2}dx$$

$$=\dfrac{\pi}{2}\int_{-1}^{3}1\cdot dx$$

$$=\dfrac{\pi}{2}[3-(-1)]$$

$$=\dfrac{\pi}{2}\times 4=2\pi$$

$$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{6}$$

$$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=tan^{-1}x\displaystyle \vert _{\frac{1}{\sqrt{3}}}^{k}$$

$$=[tan^{-1}k-tan^{\dfrac{-1}{\sqrt{3}}}]$$

$$=tan^{-1}k-tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$$

$$tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$$ ---- equation (i)

$$tan^{-1}k=\dfrac{\pi}{6}+\dfrac{\pi}{6}$$ ---- equation (ii)

$$tan^{-1}k=\dfrac{\pi}{3}$$

$$k=\sqrt{3}$$

$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9  x}{\cos ^{11}x}dx=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9  x}{\cos ^{9}x} \sec ^2  dx$$

$$\displaystyle =\int_{0}^{\dfrac{\pi}{4}} \tan ^9x d(\tan   x)$$

$$=\left[\dfrac{\tan ^{10}x}{10}+c \right]_{0}^{\dfrac{\pi}{4}}$$

$$=\left [ \left ( \dfrac{\tan ^{10}(\dfrac{\pi}{4})}{10}+c \right ) -\left ( \dfrac{\tan ^{10}(0)}{10} +c \right )\right ]$$

$$\tan \dfrac{\pi}{4}=1  ;  \tan   0=0$$

$$\Rightarrow I=\dfrac{1}{10}-\dfrac{0}{10}=\dfrac{1}{10}$$

So,  $$\displaystyle \int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9 x}{\cos ^{11}x}dx=\dfrac{1}{10}$$

We need to find value of $$\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx$$
We know that

$$\displaystyle \int   \dfrac{1}{x\sqrt{x^2-1}}dx=\sec^{-1}x+c$$

$$\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx= \sec^{-1}2-\sec^{-1}1$$

$$\sec^{-1}2 =\dfrac{\pi}{3}$$

and $$\sec^{-1}1=0$$

So,  $$\sec^{-1}2-\sec^{-1}1=\dfrac{\pi}{3}$$

$$\therefore \displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx=\dfrac{\pi}{3}$$

$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos  x}{1+sin  x}dx$$
we know that
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos  x}{1+sin  x}dx=\int_{0}^{\dfrac{\pi}{2}}\dfrac{d(sin  x)}{1+sin  x}$$
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{d (sin  x)}{1+sin  x}=log (1+sin  x)\int_{0}^{\dfrac{\pi}{2}}$$
$$=log (1+sin \dfrac{\pi}{2}) - log (1+sin  0)$$
$$=log (2)$$
$$=log 2$$

$$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2$$
$$=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2=\dfrac{1}{2}\left [ -\dfrac{1}{e^{x^2}}+c \right ] \int_{0}^{\infty}$$
$$=\dfrac{1}{2}\left [ (0+e)-(-1+e) \right ]$$
$$=\dfrac{1}{2}$$
$$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}$$

$$\displaystyle \int_{0}^{1}\dfrac{x^2dx}{1+x^2}$$

$$\displaystyle \int_{0}^{1}1{dx}-\displaystyle \int_{0}^{1}\dfrac{1}{1+x^2}dx$$

$$=1-\left [ \tan^{-1}x \right ] \displaystyle \vert_{0}^{1}$$

$$=1 - \left [ \tan^{-1}  1 \right ]$$

$$=1 -\dfrac{\pi}{4}$$

Thus $$\displaystyle \int_{0}^{1}\dfrac{x^2}{1+x^2}dx=1-\dfrac{\pi}{4}$$

$$\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=\int_{0}^{1}\dfrac{de^x}{1+e^{2x}}$$
$$=tan^{-1}(e^x)\int_{0}^{1}$$
$$=tan^{-1}(e)-tan^{-1}(e^{\circ})$$
$$=tan^{-1}(e)-tan^{-1}(1)$$
$$=tan^{-1}(e)-\dfrac{\pi}{4}$$
So,   $$\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=tan^{-1}e-\dfrac{\pi}{4}$$

$$\int_{1}^{2}\left ( \dfrac{1+x  \log  x}{x} \right ) e^x  dx$$

$$\int_{1}^{2}\left ( \dfrac{1}{x}  +  \log  x \right ) e^xdx$$           $$\displaystyle \int e^x [f(x)+f'(x) ] dx=e^x f(x)$$

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

$$=e^2  \log 2-e   \log  1$$

$$=e^2  \log  2-0$$

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

$$\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{2\sqrt{1-\dfrac{ax^2}{2}}}=\dfrac{1}{2}\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{\sqrt{1-(\dfrac{3x}{2})^2}}$$

$$=\left[\dfrac{2}{3}\times \dfrac{1}{2}  sin^{-1}\left(\dfrac{3x}{2}\right)\displaystyle \right]_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}$$

$$=\dfrac{2}{3}\times \dfrac{1}{2} \left [ sin^{-1} \left ( \dfrac{3\sqrt{3}}{3(2)} \right ) -sin^{-1} \left ( \dfrac{3\sqrt{2}}{3\times 2} \right ) \right ]$$

$$=\dfrac{1}{3} \left [ sin^{-1} \left ( \dfrac{\sqrt{3}}{2} \right ) -sin \left ( \dfrac{1}{\sqrt{2}} \right ) \right ]$$

$$=\dfrac{1}{3} \left [ \dfrac{\pi}{3}-\dfrac{\pi}{4} \right ]$$

$$=\dfrac{1}{3} \left ( \dfrac{\pi}{12} \right )$$

$$=\dfrac{\pi}{36}$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{1}{sin  x  +  cos  x }dx$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{1+tan^2 \dfrac{x}{2}}{1-tan^2\dfrac{x}{2}+2 tan \dfrac{x}{2}}  dx$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{2  d  (tan \dfrac{x}{2})}{1-(tan\dfrac{x}{2}+1)^2}=2 \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{d (tan \dfrac{x}{2})}{(\sqrt{2})^2-(tan \dfrac{x}{2}-1)^2}$$

$$=2\left [ \dfrac{1}{2\sqrt{2}} \right ]   log   \left | \dfrac{\sqrt{2}+tan\dfrac{x}{2}-1}{\sqrt{2}-(tan\dfrac{x}{2}-1)} \right | _{0}^{\dfrac{\pi}{2}}$$

$$=\dfrac{1}{\sqrt{2}} log \left ( \dfrac{\sqrt{2}+2}{\sqrt{2}-2} \right )$$

$$=log \left ( \dfrac{1+\sqrt{2}}{1-\sqrt{2}} \right )$$

$$=\sqrt{2}log (\sqrt{2}+1)$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot sin2x dx=\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot 2 sinx   cosx   dx$$

$$=2\displaystyle \int_{0}^{\dfrac{\pi}{2}}  cos^6 x   sinx  dx=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6 x\cdot d(cos  x)$$

$$=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6  x   dcos  x=-2\left ( \dfrac{cos^7x}{7} \right )\displaystyle \vert_{0}^{\dfrac{\pi}{2}}$$

$$=-2\left [ \dfrac{cos^7  x}{7}\displaystyle \vert_{0}^{\frac{\pi}{2}} \right ]$$

$$=-2\left ( \dfrac{-1}{7} \right)=\dfrac{2}{7}$$

Thus $$\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5  x\times sin 2x  dx=\dfrac{2}{7}$$

Consider, $$I=\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1-sin 2  x}{1+sin 2  x}}\cdot dx$$

$$\displaystyle I= \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{(sinx - cos x)^2}{(sin x + cos x)^2}}\cdot  dx$$

$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}{\dfrac{(sinx - cos x)}{sin x + cos x}}\cdot dx$$

$$\displaystyle I=-\int_{0}^{\dfrac{\pi}{4}}{\dfrac{cos  x-  sin  x}{sin  x +  cos  x}}dx$$

$$I=-\left[log |sin  x  +  cos  x|\right]_0^{\frac{\pi}{4}}$$

$$I=-\left[log (sin  {\frac{\pi}{4}}  +  cos  {\frac{\pi}{4}}) -\log (\sin 0+\cos 0))\right]$$

$$I=-\left[log \left (\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt 2}\right)-\log 1\right]$$

$$I=-log \left (  \dfrac{1}{\sqrt 2} +\dfrac{1}{\sqrt 2} \right ) =-log \sqrt{2}$$

So,  $$\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1- sin 2 x}{1+sin 2 x}}x=-log (\sqrt{2})$$

$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{1-x^2}}$$

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

$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{1-x^2}}=\left[(\sin^{-1}x+c) \right]_{0}^{1}$$

$$=(\sin^{-1}1+c)-(\sin^{-1}0+c)$$

$$=\dfrac{\pi}{2}$$

$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{1-x^2}}=\dfrac{\pi}{2}$$

 $$\displaystyle \int_{0}^{1}\dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx$$

$$\displaystyle \int \dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx=\displaystyle \int (sin^{-1}x)d(sin^{-1}x)$$

$$=\dfrac{(sin^{-1}x)^3}{3}+c$$

$$\displaystyle \int_{0}^{1}\dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx=\left[\dfrac{(sin^{-1}x)^3}{3}+c\right]_{0}^{1}$$

$$=\left ( \dfrac{(sin^{-1}1)^3}{3}+c\right )-\left ( \dfrac{(sin^{-1}0)^3}{3}+c\right )$$

$$=\dfrac{1}{3}\left ( \dfrac{\pi}{2} \right )^3$$

$$=\dfrac{\pi^3}{24}$$

$$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx$$

We know that $$\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx=(\sin^{-1}x+c)$$

$$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\left[(\sin^{-1}x+c)\displaystyle \right]_{\frac{1}{2}}^{1}$$

$$=(sin^{-1}1+c)-(\sin{\dfrac{1}{2}}+c)$$

$$=\sin^{-1}1-\sin^{-1}\dfrac{1}{2}$$

$$=\dfrac{\pi}{2}-\dfrac{\pi}{6}$$

$$=\dfrac{\pi}{3}$$

$$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{\pi}{3}$$

$$\int_{0}^{1}\sqrt{1-x^2}\cdot dx$$
$$\int \sqrt{1-x^2}dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2} sin^{-1}\left ( \dfrac{x}{a} \right )+ c $$
$$\int_{0}^{1}\sqrt{1-x^2}\cdot dx= \left ( \dfrac{1}{2} \sqrt{1^2-x^2}+\dfrac{a^2}{2}sin^{-1} \left ( \dfrac{x}{a} \right )+c \right )\int_{0}^{1}$$
$$=\left ( \dfrac{1}{2} (0)+ \dfrac{1}{2} sin^{-1}(1)+c \right ) - (0+0+c)$$
$$=\dfrac{1}{2}\dfrac{\pi}{2}$$
$$=\dfrac{\pi}{4}$$
$$\int_{0}^{1}\sqrt{1-x^2}dx=\dfrac{\pi}{4}$$

$$\int_{0}^{a}\dfrac{1}{a^2+x^2}dx$$

$$\int \dfrac{1}{a^2+x^2}dx=\dfrac{1}{a^2}$$

$$\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}=\dfrac{1}{a}\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}$$

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

$$\int_{0}^{a}\dfrac{1}{a^2+x^2}dx=\left [ \dfrac{1}{a} tan^{-1}\left ( \dfrac{x}{a} \right ) +c \right ] \int_{0}^{a}$$

$$=\left [ \dfrac{1}{a}tan^{-1}\left ( \dfrac{a}{a} \right ) +c \right ] -(0+c)$$
$$=\dfrac{1}{a}tan^{-1}(1)$$

$$=\dfrac{1}{a}\cdot \dfrac{\pi}{4}$$

$$=\dfrac{\pi}{4a}$$

$$\displaystyle \int_{a}^{\frac{a}{2}}\dfrac{1}{\sqrt{a^2-x^2}}\cdot dx$$

$$\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}\cdot dx=\dfrac{1}{a}\displaystyle \int \dfrac{1}{\sqrt{1-\left ( \dfrac{x}{a} \right )^2}}\cdot dx$$

$$=\displaystyle \int \dfrac{d\left ( \dfrac{x}{a} \right )}{\sqrt{1-\left ( \dfrac{x}{a} \right )^2}}$$

$$=\sin^{-1}(\dfrac{x}{a})+c$$

$$\displaystyle \int_{\frac{a}{2}}^{a}\dfrac{1}{\sqrt{a^2-x^2}}\cdot =\left [ \sin^{-1}(\dfrac{x}{a})+c \right ] _{\frac{a}{2}}^{a}$$ 

$$=(\sin^{-1}(\dfrac{a}{a})+c)-(\sin^{-1}(\dfrac{\frac{a}{2}}{a})+c)$$

$$=\left [ (\sin (1)+c ) -(\sin^{-1}(\dfrac{1}{2})+c) \right ]$$

$$=\dfrac{\pi}{2}-\dfrac{\pi}{6}$$

$$=\dfrac{\pi}{3}$$

$$\displaystyle \int_{0}^{1}\dfrac{x}{a+x^2}dx$$

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

$$=\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\left [ \dfrac{1}{2} \log (1+x^2)+c \right ]_{0}^{1}$$

$$=\left ( \dfrac{1}{2} \log (2)+c \right ) - \left ( \dfrac{1}{2} \log (1)+c\right )$$

$$=\dfrac{1}{2} \log 2$$

$$\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\dfrac{1}{2} \log 2$$

Let $$\displaystyle I=\int_{0}^{1}\dfrac{x^3}{1+x^8} dx$$

Now, $$\displaystyle \int \dfrac{x^3}{1+x^8 }dx =\dfrac{1}{4}\int \dfrac{d(x^4)}{1+x^8}=\dfrac{1}{4}\tan^{-1}(x^4)+c$$

$$I=\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx=\left(\dfrac{1}{4}\tan^{-1}(x^4)+c\right)_{0}^{1}$$

$$=\left [\dfrac{1}{4} \tan^{-1}(1)+c\right ]-\left [\dfrac{1}{4} \tan^{-1}(0)+c\right ]$$

$$=\dfrac{1}{4} \tan^{-1}(1)$$

$$=\dfrac{1}{4} \cdot \dfrac{\pi}{4}=\dfrac{\pi}{16}$$

Hence, $$\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx =\dfrac{\pi}{16}$$