\(\Gamma \)(n) =\(\int _{ 0 }^{ \infty }{ { t }^{ n-1 }e^{ -t } } \) என்பதை நாம் அறிவோம்.
இதில் t=x² என dt=2xdx ஆகும்.
∴ \(\Gamma \)(n) =\(\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }({ x }^{ 2 })^{ n-1 } } 2xdx\)
=\(\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }x^{ 2x-2 } } 2xdx\)
=\(2\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }x^{ 2x-1 } } dx\)
n=\(\frac{1}{2}\) என, நாம் பெறுவது
\(\Gamma \left( \frac { 1 }{ 2 } \right) =2\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } } } dx\)
⇒ \(\sqrt { \pi } \int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } } } dx\)
∴ \(\int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } } } dx=\frac { \sqrt { \pi } }{ 2 } \)

\(\int { { \sin }^{ 2 }xdx } = \int { \frac { 1 }{ 2 } \left( 1-\cos2x \right) dx } \)
\(=\frac { 1 }{ 2 } \left[ \int { dx-\int { \cos2xdx } } \right] \)
\(=\frac { 1 }{ 2 } \left[ x-\frac { \sin2x }{ 2 } \right] +c\)
எளியச் சார்புகளாக பிரித்தெழுத
cos2x = 1− 2sin² x 
\(\therefore \sin^{ 2 }x=\frac { 1 }{ 2 } (1-\cos2x)\)

\(\int \sqrt { 3x+5 } dx=\int (3x+5)^{ \frac { 1 }{ 2 } }dx\)
\(=\frac { (3x+5)^{ \frac { 1 }{ 2 } +1 } }{ 3\left( \frac { 1 }{ 2 } +1 \right) } +c\)
\(=\frac { (3x+5)^{ \frac { 3 }{ 2 } } }{ 9/2 } +c\)
\(=\frac { 2 }{ 9 } (3x+5)^{ \frac { 3 }{ 2 } }+c\)
 

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

\(\int { \frac { { x }^{ 3 }dx }{ \sqrt { x^{ 8 }+1 } } } =\frac { 1 }{ 4 } \int { \frac { 4x^{ 3 }dx }{ \sqrt { (x^{ 4 })^{ 2 }+1^{ 2 } } } } \)
=\(\frac { 1 }{ 4 } \log|{ x }^{ 4 }+\sqrt { ({ x }^{ 4 })^{ 2 }+1^{ 2 } } |+c\)
=\(\frac { 1 }{ 4 } \log|{ x }^{ 4 }+\sqrt { { x }^{ 8 }+1 } |+c\).

\(\int \left( e^{ x\log a }+c^{ a\log a }-e^{ n\log x } \right) dx\)
\(=\int \left( e^{ \log a^{ x } }+c^{ \log a^{ a } }-e^{ \log x^{ n } } \right) dx\)
\(=\int \left( a^{ x }+a^{ a }-x^{ n } \right) dx\)
\(=\frac { a^{ x } }{ \log a } +a^{ a }x-\frac { x^{ n+1 } }{ n+1 } +c\)

\(\int (x\ logx) dx \)
u = logx             
du = \(\frac { 1 }{ x } \) dx
\(\int dv\)=\(\int x ^n dx \) 
v =\(\frac { x ^ n+1 }{ n +1 } \)
\(\int udv=uv-\int vdu\)
 \(\int x^{ n }\) \(\log xdx\)     =  \(\frac { x^{ n+1 } }{ n+1 } \log x-\int \frac { x^{ n+1 } }{ n+1 } \cdot \frac { 1 }{ x } dx\)
\(=\frac { x^{ n+1 } }{ n+1 } \log x-\frac { 1 }{ n+1 } \int x^{ n }dx\)
\(=\frac { x^{ n+1 } }{ n+1 } \log x-\frac { 1 }{ n+1 } \left( \frac { x^{ n+1 } }{ n+1 } \right) +c\)
\(=\frac { x^{ n+1 } }{ n+1 } \left[ \log x-\frac { 1 }{ n+1 } \right] +c\)

\(\int { { \left( \log x \right) }^{ 2 } } dx= \int { udv } \)
= uv − \(\int { } \)vdu
= x (log x)² − 2\(\int { } \) logxdx...(*)
\(=x(\log x)^{ 2 }-2\int { udv } \)
\(=x(\log x)^{ 2 }-2[uv-\int { udv } ]\)
\(=x(logx)^{ 2 }-2[xlogx-\int { dx] } \)
\(=x(logx)^{ 2 }-2xlogx+x+c\)
\(=x[(log{ ) }^{ 2 }-log{ x }^{ 2 }+2]+c\)

\(\int _{ 2 }^{ 3 }{ f(5-x) } dx-\int _{ 2 }^{ 3 }{ f(x) } dx\)
\(\int _{ 2 }^{ 3 }{ f(5-x) } dx-\int _{ 2 }^{ 3 }{ f(2+3-x) } dx\)
\(​​\left[∵ \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a+b-x)dx } \right] \)

\(\int _{ 0 }^{ - }{ { e }^{ -ex } } dx= \left [\frac {e ^ {-2x }}{-2}\right ]^ \infty _ {0}\)
\(=-\frac {1}{2} [e ^ {-\infty} - e ^0] =\frac {-1}{2} [0-1] =\frac {1}{2} \)

t = log x என்க.\(\Rightarrow dt = \frac { 1 }{ x } dx\)
\(\int \frac { \log x }{ x } dx =\int tdt = { \frac { { t }^{ 2 } }{ 2 } } +c = \frac {1}{2}(\log x ) ^ 2+c \)

\(\\ \int { \frac { sin5x-sinx }{ cos3x } } \)dx = \(\int { \frac { 2\sin \frac {5x-x }{2}\cos \frac {5x+x }{2} }{ \cos3x } } \)
=2 ∫sin 2x dx = \(2\left [\frac {−\cos 2x}{2} \right] +c \)
= −cos2x + c 

\(\int { \frac { 1 }{ { x }^{ 3 } } dx } =\int { { { x }^{ 3 } } dx } = \frac { x^ {-2} }{-2} + c = \frac { -1 }{ 2x^{ 2 } } +x\)