Explanation
∫e31dxx√1+lnx=∫e31dlnx√1+lnx
∫e31dlnx√1+lnx=2√1+lnx|e31
=2√1+3−2√1
=2√4−2=2(1)
=2
So, ∫e31dxx√1+lnx=2
We have,
I=∫10tan−1x1+x2dx
Let t=tan−1x
dt=dx1+x2
Therefore,
I=∫π40tdt
I=[t22]π40
I=12((π4)2−0)
I=12×π216
I=π232
Hence, this is the answer.
∫π30cosx3+4sinxdx=∫π30d(sinx)3+4sinx
=∫π30d(sinx)3+4sinx=14[log(3+4sinx)]π30
=14[log[3+4(sinπ3)]−log(3+4sin(0))]
=14(log(3+4√32)−log(3))
=14(log(3+2√3)−log3)
=14log(3+2√33)
∴∫π30cosx3+4sinxdx=14log(3+2√33)
∫π20cosx1+sin2xdx=∫π20dsinx1+sin2x
∫π20dsinx1+sin2x=[(tan−1(sinx)+c)]π20 (∫11+x2=tan−1x)
=(tan−1(sin(π2)+c)−(tan−1(sin0)+c)
=tan−1(1)
=π4
We need to find the value of ∫3−1[tan−1(xx2+1)+tan−1(x2+1x)]dx
We know tan−1p+cot−1p=π2
⇒tan−1p+tan−11p=π2
So, tan−1(xx2+1)+tan−1(x2+1x)=π2
∫3−1[tan−1(x1+x2)+tan−1(x2+1x)]dx
=∫3−1π2dx
=π2∫3−11⋅dx
=π2[3−(−1)]
=π2×4=2π
∫k1√311+x2dx=π6
∫k1√311+x2dx=tan−1x|k1√3
=[tan−1k−tan−1√3]
=tan−1k−tan−11√3=π6
tan−11√3=π6 ---- equation (i)
tan−1k=π6+π6 ---- equation (ii)
tan−1k=π3
k=√3
I=∫π40sin9xcos11xdx=∫π40sin9xcos9xsec2dx
=∫π40tan9xd(tanx)
=[tan10x10+c]π40
=[(tan10(π4)10+c)−(tan10(0)10+c)]
tanπ4=1;tan0=0
⇒I=110−010=110
So, ∫π40sin9xcos11xdx=110
We need to find value of ∫211x√x2−1dxWe know that
∫1x√x2−1dx=sec−1x+c
∫211x√x2−1dx=sec−12−sec−11
sec−12=π3
and sec−11=0
So, sec−12−sec−11=π3
∴∫211x√x2−1dx=π3
∫π20cosx1+sinxdx we know that ∫π20cosx1+sinxdx=∫π20d(sinx)1+sinx ∫π20d(sinx)1+sinx=log(1+sinx)∫π20 =log(1+sinπ2)−log(1+sin0) =log(2) =log2
∫∞0xe−x2dx=12∫∞0e−x2dx2 =12∫∞0e−x2dx2=12[−1ex2+c]∫∞0 =12[(0+e)−(−1+e)] =12 ∫∞0xe−x2dx=12
∫10x2dx1+x2
∫101dx−∫1011+x2dx
=1−[tan−1x]|10
=1−[tan−11]
=1−π4
Thus ∫10x21+x2dx=1−π4
∫10ex1+e2xdx=∫10dex1+e2x =tan−1(ex)∫10 =tan−1(e)−tan−1(e∘) =tan−1(e)−tan−1(1) =tan−1(e)−π4 So, ∫10ex1+e2xdx=tan−1e−π4
∫21(1+xlogxx)exdx ∫21(1x+logx)exdx ∫ex[f(x)+f′(x)]dx=exf(x) =∫21(exlogx+c) =e2log2−elog1 =e2log2−0 ∫21(1+xlogxx)exdx=e2log2
∫√33√23dx2√1−ax22=12∫√33√23dx√1−(3x2)2
=[23×12sin−1(3x2)]√33√23
=23×12[sin−1(3√33(2))−sin−1(3√23×2)]
=13[sin−1(√32)−sin(1√2)]
=13[π3−π4]
=13(π12)
=π36
∫π201sinx+cosxdx
∫π201+tan2x21−tan2x2+2tanx2dx
∫π202d(tanx2)1−(tanx2+1)2=2∫π20d(tanx2)(√2)2−(tanx2−1)2
=2[12√2]log|√2+tanx2−1√2−(tanx2−1)|π20
=1√2log(√2+2√2−2)
=log(1+√21−√2)
=√2log(√2+1)
∫π20cos5x⋅sin2xdx=∫π20cos5x⋅2sinxcosxdx
=2∫π20cos6xsinxdx=−2∫π20cos6x⋅d(cosx)
=−2∫π20cos6xdcosx=−2(cos7x7)|π20
=−2[cos7x7|π20]
=−2(−17)=27
Thus ∫π20cos5x×sin2xdx=27
∫a2a1√a2−x2⋅dx
∫1√a2−x2⋅dx=1a∫1√1−(xa)2⋅dx
=∫d(xa)√1−(xa)2
=sin−1(xa)+c
∫aa21√a2−x2⋅=[sin−1(xa)+c]aa2
=(\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}
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}
\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}
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