\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+\sqrt { tanx } } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cosx } dx }{ \sqrt { cosx } +\sqrt { sinx } } } \) ....(i)
Apply the property
\(\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x)dx, } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cos\left( \frac { \pi }{ 2 } -x \right) } }{ \sqrt { cos\left( \frac { \pi }{ 2 } -x \right) +\sqrt { sin\left( \frac { \pi }{ 2 } -x \right) } } } } \)
\(\Rightarrow I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { sinx } } } dx\) ..(ii)
by adding eqn. (i) and (ii),
\(\Rightarrow 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cosx } +\sqrt { sinx } }{ \sqrt { cosx } +\sqrt { sinx } } } dx\)
\(\Rightarrow 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } \Rightarrow \quad \left[ x \right] ^{ \frac { \pi }{ 2 } }_{ 0 }\)
\(\Rightarrow 2I=\left( \frac { \pi }{ 2 } -0 \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow I=\frac { \pi }{ 4 } \)

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

\(\int { \frac { { e }^{ x } }{ x } } dx-\int { \frac { e^{ x } }{ x^{ 2 } } } dx\)
\(=\frac { 1 }{ x } \int { e^{ x }-\int { \left( \frac { d }{ dx } \left( \frac { 1 }{ x } \right) \int { { e }^{ x }dx } \right) } } dx-\int { \frac { { e }^{ x } }{ x^{ 2 } } } \)
\(=\frac { { e }^{ x } }{ x } -\int { -\frac { 1 }{ { x }^{ 2 } } { e }^{ x }dx } -\int { \frac { { e }^{ x } }{ { x }^{ 2 } } } dx\)
\(=\frac { { e }^{ x } }{ x } +\int { \frac { { e }^{ x } }{ x^{ 2 } } dx } -\int { \frac { { e }^{ x } }{ { x }^{ 2 } } } dx\)
\(=\frac { { e }^{ x } }{ x } +C\)

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

\(I=\int { { e }^{ x }(secxtanx+secx)dx } \)
\(={ e }^{ x }secx+C\)
\([\because \int { { e }^{ x }\left\{ f(x)+f'(x) \right\} } dx={ e }^{ x }f(x)+C\)
 \(\therefore \quad f(x)=secx\)

\(={2\over3}(x+2)^{3/2}-4\sqrt{x+2}+c\)

\(= -{1\over3}cot^3\ x+c\)

\({(ax +b)^4\over 4a}+c\)​​​​​​​

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

\(I=\int _{ 2 }^{ 5 }{ \left[ \left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| \right] } dx\)
\(=\int _{ 2 }^{ 3 }{ f(x)dx+\int _{ 3 }^{ 5 }{ f(x)dx, } } \)
where \(f(x)=\left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| \)
 =\(\{\)\(x-2-x+3-x+5=6-x,\quad if\quad 2\le x\le 3\\ x-2+x-3-x+5=x,\quad if\quad 3\le x\ge 5\)
Now, from (i) we have 
\(I=\int _{ 2 }^{ 3 }{ (6-x)dx+ } \int _{ 3 }^{ 5 }{ xdx } \)
\(=\left[ 6x-\frac { { x }^{ 2 } }{ 2 } \right] ^{ 3 }_{ 2 }+\left[ \frac { { x }^{ 2 } }{ 2 } \right] ^{ 5 }_{ 3 }\)
\(=\left( 18-\frac { 9 }{ 2 } \right) -(12-2)+\frac { 1 }{ 2 } (25-9)\)
\(=\frac { 23 }{ 2 } \)

\(I=\int { \frac { 3x+1 }{ (x+1)^{ 2 }(x+3) } } dx\)
We resolve the integrand into partial fractions
Let \(\int { \frac { 3x+1 }{ (x+1)^{ 2 }(x+3) } } =\frac { A }{ x+1 } +\frac { B }{ (x+1)^{ 2 } } +\frac { C }{ x+3 } \)
\(3x+1=A(x+1)(x+3)+B(x+3)+C(x+1)^{ 2 }\)
Solving, we get A = 2, B = -1, C = -2
From(i),
\(\int { \frac { 3x+1 }{ (x+1)^{ 2 }(x+3) } } =\frac { 2 }{ x+1 } -\frac { 1 }{ (x+1)^{ 2 } } -\frac { 2 }{ x+3 } \)
\(I=\int { \frac { 2 }{ x+1 } dx-\int { \frac { 1 }{ (x+1)^{ 2 } } } dx-\int { \frac { 2 }{ x+3 } } dx } \)
\(=2log\left| x+1 \right| +\frac { 1 }{ x+1 } -2log\left| x+3 \right| +C\)
\(=2log\frac { (x+1) }{ (x+3) } +\frac { 1 }{ x+1 } +C\)

\(\int { \frac { (x+3) }{ { (x+5) }^{ 3 } } } { e }^{ x }dx=\int { \frac { x+5-1 }{ { (x+5) }^{ 3 } } } { e }^{ x }dx\)
\(=\int { \left[ \frac { 1 }{ { (x+5) }^{ 2 } } -\frac { 2 }{ { (x+5) }^{ 3 } } \right] } { e }^{ x }dx\)
\(=\int { \frac { 1 }{ { (x+5) }^{ 2 } } } { e }^{ x }-\int { \frac { 2 }{ { (x+5) }^{ 3 } } } { e }^{ x }dx\)
\(=\frac { 1 }{ { (x+5) }^{ 2 } } { e }^{ x }-\int { \frac { -2 }{ { (x+5) }^{ 3 } } } { e }^{ x }dx-\int { \frac { 2 }{ { (x+5) }^{ 3 } } } { e }^{ x }dx\)
[Integrating first integral by Parts]
\(=\frac { { e }^{ x } }{ { (x+5) }^{ 2 } } +c\)

\(I=\int { \frac { { \left( \log { x } \right) }^{ 2 } }{ x } } dx.\)
\(Put\ \log { x } =t\)  so that  \(\frac { 1 }{ x } dx=dt\)
\(\therefore \ I=\int { { t }^{ 2 } } dt=\frac { { t }^{ 3 } }{ 3 } +c\)
\(=\frac { { \left( \log { x } \right) }^{ 2 } }{ 3 } +C,\ x>0.\)

Put \(\sin { x } -\cos { x=t } \) so that \((\cos { x } +\sin { x } )=dt.\)
Squaring \(\sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } -2\sin { x } \cos { x } ={ t }^{ 2 }\)
\(\Rightarrow 1-\sin { 2x } ={ t }^{ 2 } \Rightarrow \sin { 2x } =1-{ t }^{ 2 }.\)
When \(x=\frac { \pi }{ 6 } , t=\sin { \frac { \pi }{ 6 } } -\cos { \frac { \pi }{ 6 } } =\frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } .\)
When  \(x=\frac { \pi }{ 3 } , t=\sin { \frac { \pi }{ 3 } } -\cos { \frac { \pi }{ 3 } } =\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } .\)
\(\therefore \ I= \int _{ \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } }^{ \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } }{ \frac { dt }{ \sqrt { t-{ t }^{ 2 } } } } ={ \left[ \sin ^{ -1 }{ t } \right] }_{ \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } }^{ \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } }\)
\(=\sin ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } \right) } -\sin ^{ -1 }{ \left( \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \right) } \)
\(=\sin ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } \right) } \)
\(=2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \left( \sqrt { 3 } -1 \right) .\)

\(\int\left(a x^{2}+b x+c\right) d x\)
\(=a \int x^{2} d x+b \int x d x+c \int 1 \cdot d x \)
\(=a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)+c x+C \)
\(=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+C \)

\(\text { Let } \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x\)
Then, by P₄
\(\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\)
\(\text { Hence } \ \mathrm{I}=\frac{\pi}{4}\)

The integrand is a proper rational function. Therefore, by using the form of partial fraction. we write
\(\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\)
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A + B = 0
and 2A + B = 1
Solving these equations, we get A = 1 and B = – 1.
Thus, the integrand is given by
\(\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}\)
\(\text { Therefore, }\int \frac{d x}{(x+1)(x+2)}=\int \frac{d x}{x+1}-\int \frac{d x}{x+2} \)
\(=\log |x+1|-\log |x+2|+\mathrm{C}\)
\(=\log \left|\frac{x+1}{x+2}\right|+\mathrm{C}\)

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

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { x } } }{ \sqrt { \sin { x } } +\sqrt { \cos { x } } } } dx\)
Apply property \(\int _{ a }^{ b }{ f\left( x \right) } =\int _{ a }^{ b }{ f\left( a+b-x \right) } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } }{ \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } +\sqrt { \cos { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } } } dx\)
\(\therefore I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \cos { x } } }{ \sqrt { \cos { x } } +\sqrt { \sin { x } } } } dx\)
Adding, we get \(2I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)

Let, \(I=\int { (5x-1)\sqrt { 6+5x-2x^{ 2 } } } dx\)
Take, \(5x-1=A\frac { d }{ dx } (6+5x-2x^{ 2 })+B\)
\(=A(5-4x)+B\)
\(\Rightarrow 5=-4a \ \text{and} -1=5A+B\)
\(\Rightarrow A=-\frac { 5 }{ 4 } , B=\frac { 21 }{ 4 } \)
\(\therefore I=-\frac { 5 }{ 4 } \int { (5-4x) } \sqrt { 6+5x-2x^{ 2 } } dx+\frac { 21 }{ 4 } \int { \sqrt { 2 } } \sqrt { 3+\frac { 5 }{ 2 } x-{ x }^{ 2 } } dx\)
\(=-\frac { 5 }{ 4 } \times \frac { 2 }{ 3 } (6+5x-2x^{ 2 })^{ 3/2 }+\frac { 21\sqrt { 2 } }{ 4 } \int { \sqrt { \left( \frac { \sqrt { 73 } }{ 4 } \right) ^{ 2 }-\left( x-\frac { 5 }{ 4 } \right) ^{ 2 } } } dx\)
\(=-\frac { 5 }{ 6 } (6+5x-2x^{ 2 })^{ 3/2 }+\frac { 21\sqrt { 2 } }{ 4 } \left[ \frac { x-\frac { 5 }{ 4 } }{ 2 } \sqrt { 3+\frac { 5 }{ 2 } x-x^{ 2 } } +\frac { 73 }{ 2\times 16 } { sin }^{ -1 }\left( \frac { x-\frac { 5 }{ 4 } }{ \frac { \sqrt { 73 } }{ 4 } } \right) \right] +C\)
\(=-\frac { 5 }{ 6 } (6+5x-2x^{ 2 })^{ 3/2 }+\frac { 21\sqrt { 2 } }{ 4 } \left[ \frac { 4x-5 }{ 8 } \sqrt { 3+\frac { 5 }{ 2 } x-{ x }^{ 2 } } +\frac { 73 }{ 32 } { sin }^{ -1 }\left( \frac { 4x-5 }{ \sqrt { 73 } } \right) \right] +C\)