\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ sinx } }{ \sqrt [ 3 ]{ cosx } +\sqrt [ 3 ]{ sinx } } } dx\)
Apply the property
\(\int _{ 0 }^{ b }{ f(x)\quad dx } =\int { f(a+b-x)dx } \)
\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ sin\left( \frac { \pi }{ 2 } -x \right) dx } }{ \sqrt [ 3 ]{ cos\left( \frac { \pi }{ 2 } -x \right) } +\sqrt [ 3 ]{ sin\left( \frac { \pi }{ 2 } -x \right) } } } \)
as   \(\frac { \pi }{ 3 } +\frac { \pi }{ 6 } =\frac { 2\pi +\pi }{ 6 } =\frac { 3\pi }{ 6 } =\frac { \pi }{ 2 } \)
\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ cosx } }{ \sqrt [ 3 ]{ cosx } +\sqrt [ 2 ]{ sinx } } } dx\)
by adding eqn. (i) and (ii),
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] ^{ \pi /3 }_{ \pi /6 }\)
\(=\left[ \frac { \pi }{ 3 } -\frac { \pi }{ 6 } \right] =\frac { \pi }{ 6 } \)
\(\Rightarrow 2I=\frac { \pi }{ 6 }\)
\( \Rightarrow I=\frac { \pi }{ 12 } \)

\( \int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1 \)
\( =\tan ^{-1} 1-\tan ^{-1} 0 \\ \)
\( =\frac{\pi}{4}-0=\frac{\pi}{4} \)

\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx } dx=\left[ log\left| secx \right| \right] ^{ \pi /4 }_{ 0 }\)
\(=log\left| sec\frac { \pi }{ 4 } \right| -log\left| sec0 \right| \)
\(=log\left| \sqrt { 2 } \right| -log\left| 1 \right| \)
\(\therefore \ \int _{ 0 }^{ \pi /4 }{ tanxdx } =\frac { 1 }{ 2 } log2\)

\(I=\int _{ 1 }^{ 2 }{ \left( \frac { { x }^{ 3 } }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } \right) } dx\)
\(=\int _{ 1 }^{ 2 }{ \left( x-\frac { 1 }{ { x }^{ 2 } } \right) dx } \)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } +\frac { 1 }{ x } \right] ^{ 2 }_{ 1 }\)
\(I=\left( 2+\frac { 1 }{ 2 } \right) -\left( \frac { 1 }{ 2 } +1 \right) \)
\(=\frac { 5 }{ 2 } -\frac { 3 }{ 2 } \)
\(\Rightarrow I=1\)

\(\int { { e }^{ x }cotx } dx+\int { { e }^{ x }(logsinx) } dx\)
\(=\int { { e }^{ x } } cotxdx+log(sinx)\int { { e }^{ x }dx } -\int { \left[ \frac { d }{ dx } (logsinx)\int { { e }^{ x }dx } \right] } \)
\(=\int { { e }^{ x }cotx } dx-{ e }^{ x }log(sinx)-\int { cotx{ e }^{ x }dx } \)
\(={ e }^{ x }log\left| sinx \right| +C\)

\(\int { \frac { (x+1){ e }^{ x } }{ cos^{ 2 }(xe^{ x }) } } dx\)
Put,\(x{ e }^{ x }=t\)
\(\Rightarrow (xe^{ x }+{ e }^{ x })dx=dt\)
\({ \Rightarrow e }^{ x }(x+1)dx=dt\)
\(\int { \frac { dt }{ cos^{ 2 }t } } =\int { sec^{ 2 }tdt } \)
\(=tant=tan(xe^{ x })+C\)

\(\int { tan^{ -1 } } \sqrt { \frac { 1-cos2x }{ 1+cos2x } } dx=\int { tan^{ -1 } } \sqrt { \frac { 2sin^{ 2 }x }{ 2cos^{ 2 }x } } \)
\(=\int { { tan }^{ -1 }(tanx) } dx\)
\(=\int { x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } +C\)

\(cosC.sinD=\frac { 1 }{ 2 } \left[ sin(C+D)-sin(C-D) \right] \)
\(=\frac { 1 }{ 2 } \int { \left[ sin(3x+2x)-sin(3x-2x) \right] } dx\)
\(=\frac { 1 }{ 2 } \int { (sin5x-sinx) } dx\)
\(=\frac { 1 }{ 2 } \left( -\frac { cos5x }{ 5 } \right) -\frac { 1 }{ 2 } \left[ -cosx \right] \)
\(=\frac { 1 }{ 2 } \left[ cosx-\frac { 1 }{ 5 } cos5x \right] +C\)

\(\int { { cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 2 } -x \right) \right] dx } =\int { \left( \frac { \pi }{ 2 } -x \right) } dx\)
\(=\int { \frac { \pi }{ 2 } } dx-\int { xdx } \)
\(=\frac { \pi }{ 2 } x-\frac { { x }^{ 2 } }{ 2 } +c\)

\(= {\pi\over2}\times-{x^2\over 2}+c\)

Consider : \(\int({cos\ x-sin\ x})^2dx=\int({cos^2x + sin^2\ x-2sin\ x\ \ cos\ x})dx \)
\(=\int({1-sin\ 2\ x})dx=x+{cos\ 2x\over 2}+ C\)

\(= log\ |{1-cos\ x\over 2-cos\ x}|+\ c\)

= tan x + sec x + c

\(\int \frac{\sec ^{2} x}{3+\tan x} d x =\int \frac{1}{t} d t \)
= log |t|+C
= log | 3 + tan x | + C

\(\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x=\frac{1}{6} \int \frac{1}{t} d t=\frac{1}{6} \log |t|+C \)
\(={1\over6}log|3x^2 + sin6x|+C|\)

\(=-{2\over3}cosec\quad x+C\)

\(=\int { \left( { x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } } \right) } dx\)
\(=\int { { x }^{ \frac { 1 }{ 2 } } } dx-\int { { x }^{ \frac { 3 }{ 2 } } } dx\)
\(=\frac { { x }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } -\frac { { x }^{ \frac { 3 }{ 2 } +1 } }{ \frac { 3 }{ 2 } +1 } +C\)
\(=\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+C.\)

To start with, we are unable to guess a function whose derivative is log x. We take log x as the first function and the constant function 1 as the second function. Then, the integral of the second function is x
\(\text { Hence, } \int(\log x .1) d x =\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x \)
\(=(\log x) \cdot x-\int \frac{1}{x} x d x=x \log x-x+\mathrm{C} . \)

The integrand is a proper rational function. Decompose the rational function into partial fraction. Write
\(\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{A}{x+2}+\frac{B x+C}{\left(x^{2}+1\right)}\)
Therefore, x² + x + 1 = A (x² + 1) + (Bx + C) (x + 2)
Equating the coefficients of x², x and of constant term of both sides, we get A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get
\(\mathrm{A}=\frac{3}{5}, \mathrm{~B}=\frac{2}{5} \text { and } \mathrm{C}=\frac{1}{5}\)
Thus, the integrand is given by
\(\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{x^{2}+1}=\frac{3}{5(x+2)}+\frac{1}{5}\left(\frac{2 x+1}{x^{2}+1}\right)\)
\(\text { Therefore, } \int \frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)} d x=\frac{3}{5} \int \frac{d x}{x+2}+\frac{1}{5} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{5} \int \frac{1}{x^{2}+1} d x\)
\(=\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+\mathrm{C}\)

We have :\(f\left( x \right) =\int _{ 0 }^{ x }{ tsin } t\quad dt\)
ஃ \(f^{ \prime }\left( x \right) =x\sin { \frac { d }{ dx } (x) } -0=x(1)=x\sin { x } .\)

Put \({ e }^{ x }\)= t so that \({ e }^{ x }\) dx = dt.
When x = 0, t = \({ e }^{ 0 }\)= 1. When x = 1, t = e.
ஃ \(I=\int _{ 1 }^{ e }{ \frac { dt }{ 1+{ t }^{ 2 } } = } { \left[ \tan ^{ -1 }{ t } \right] }_{ 1 }^{ e }\)
\(={ tan }^{ -1 } e-\tan ^{ -1 }{ 1 } =\tan ^{ -1 }{ e-\frac { \pi }{ 4 } } \)

\({1\over2\sqrt2}tan^{-1}({x^2-4\over2\sqrt2 x})+c\)

\(\int { \frac { sin^{ -1 }\sqrt { x } -cos^{ -1 }\sqrt { x } }{ sin^{ -1 }\sqrt { x } +cos^{ -1 }\sqrt { x } } } dx\)
\(=\frac { 2 }{ \pi } \int { \left[ sin^{ -1 }\sqrt { x } -\left( \frac { \pi }{ 2 } -{ sin }^{ -1 }\sqrt { x } \right) \right] } dx\)
\(=\frac { 2 }{ \pi } \int { 2{ sin }^{ -1 }\sqrt { x } } dx-\int { 1dx } \)
\(=\frac { 4 }{ \pi } \left[ { sin }^{ -1 }\sqrt { x } .x-\int { \frac { 1 }{ \sqrt { 1-x } } \frac { 1 }{ 2\sqrt { x } } .xdx } \right] -x+C\)
\(=\frac { 1 }{ \pi } \left[ 4x{ sin }^{ -1 }\sqrt { x } -2\int { \frac { \sqrt { x } }{ \sqrt { 1-x } } } dx \right] -x+C\)
Let, \(x={ sin }^{ 2 }\theta \Rightarrow dx=2sin\theta cos\theta \quad d\theta \)
\(=\frac { 1 }{ \pi } \left[ 4x{ sin }^{ -1 }\sqrt { x } -2\int { \frac { sin\theta }{ cos\theta } 2sin\theta cos\theta \quad d\theta } \right] \)
\(=\frac { 1 }{ \pi } \left[ 4x{ sin }^{ -1 }\sqrt { x } -2\int { (1-cos2\theta )d\theta } \right] -x+C\)
\(=\frac { 1 }{ \pi } \left[ 4x{ sin }^{ -1 }\sqrt { x } -2\left( \theta -\frac { sin2\theta }{ 2 } \right) \right] -x+C\)
\(=\frac { 1 }{ \pi } \left[ 4x{ sin }^{ -1 }\sqrt { x } -2{ sin }^{ -1 }\sqrt { x } +2\sqrt { x } \sqrt { 1-x } \right] -x+C\)
\(=\frac { sin^{ -1 }\sqrt { x } }{ \pi } 2(2x-1)+\frac { 2 }{ \pi } \sqrt { x-{ x }^{ 2 } } -x+C\)