Trigonometry> Find the integration of that two question...

Find the integration of that two questions according to the respective formulas

Ankith , 7 Years ago

Grade 11

1 Answers

Sami Ullah

Last Activity: 7 Years ago

Question:- $\int cos^{11}x.sin^{3}x.dx$

$I=\int cos^{11}x.(1-cos^{2}x)sinx.dx$

$I=\int -cos^{11}x.(cos^{2}x-1)sinx.dx$

Subtitute $u=cosx$ $\rightarrow$ $du=-sinx.dx$

So we get,

$I=\int u^{11}.(u^{2}-1).du$

$I=\int( u^{13}-u^{11}).du$

$I=\int( u^{13}).du-\int (u^{11}).du$

$I=\frac{u^{14}}{14}-\frac{u^{12}}{12}+C$

$I=\frac{cos^{14}x}{14}-\frac{cos^{12}x}{12}+C$

Question 2:- $I=\int sin^{4}\Theta .cos^{2}\Theta .d\Theta$

$I=\int sin^{4}\Theta .(1-sin^{2}\Theta) .d\Theta$

$I=\int( sin^{4}\Theta-sin^{6}\Theta) .d\Theta$

$I=\int( sin^{4}\Theta).d\Theta- \int (sin^{6}\Theta) .d\Theta$

Now we reduce the above by reduction formula and then integrate them.

Reduction formula is,

$\int sin^{n}\Theta .d\Theta =\frac{n}{n-1}\int (sin^{n-2}\Theta).d\Theta-\frac{cos\Theta.sin^{n-1}\Theta }{n}$

Put n=4 and n=6 respectively.Thus we get

$I=\frac{4}{4-1}\int (sin^{4-2}\Theta).d\Theta-\frac{cos\Theta.sin^{4-1}\Theta }{4}-\frac{6}{6-1}\int (sin^{6-2}\Theta).d\Theta-\frac{cos\Theta.sin^{6-1}\Theta }{6}$

$I=\frac{4}{3}\int (sin^{2}\Theta).d\Theta-\frac{cos\Theta.sin^{3}\Theta }{4}-\frac{6}{5}\int (sin^{4}\Theta).d\Theta+\frac{cos\Theta.sin^{5}\Theta }{6}$

Now you can simply integrate the reamining $sin^2\Theta$ and $sin^4\Theta$ by the reduction formula.Also leave the the terms as they are which have no integral sign with them.