Mathematics

Evaluate : $$\underset{0}{\overset{\frac{\pi}{2}}{\int}} \sin^4x \times \cos^5x. dx$$


SOLUTION
$$\int_{0}^{\frac{\pi}{2}}\sin ^4x\cos ^5xdx$$

$$=\int_{0}^{\frac{\pi}{2}}\sin ^4x(\cos ^2x)^2\cos xdx$$

$$=\int_{0}^{\frac{\pi}{2}}\sin ^4x(1-\sin ^2x)^2\cos xdx$$

substitute $$u=\sin x\Rightarrow du=\cos x dx$$

$$=\int_{0}^{1}u^4(1-u^2)^2du$$

$$=\int_{0}^{1}u^4-2u^6+u^8du$$

$$=\left [ \dfrac{u^5}{5}-2\dfrac{u^7}{7}+\dfrac{u^9}{9} \right ]_0^1$$

$$=\left [ \dfrac{\sin ^5x}{5}-2\dfrac{\sin ^7x}{7}+\dfrac{\sin ^9x}{9} \right ]_0^1+C$$

$$=\dfrac{8}{315}-0$$

$$\int_{0}^{\frac{\pi}{2}}\sin ^4x\cos ^5xdx=\dfrac{8}{315}$$
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Subjective Medium Published on 17th 09, 2020
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