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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 19

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 19

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Answer:   \frac{2}{35}-\frac{9}{280 \sqrt{2}}

Given:   \int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x

Hint: Use trigonometric identities.

Solution:   \int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin x\left(1-\cos ^{2} x\right) d x \\ & \end{aligned}

\left(\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \sin ^{2} x=1-\cos ^{2} x \end{array}\right)

\begin{aligned} &\text { Let } \\ & \end{aligned}

\cos x=t \\                          (Differentiate w.r.t to x)

-\sin x d x=d t

\begin{aligned} &=-\int_{0}^{\frac{\pi}{4}} t^{4}\left(1-t^{2}\right) d t \\ & \end{aligned}

=\int_{0}^{\frac{\pi}{4}} t^{4}\left(t^{2}-1\right) d t \\

=\int_{0}^{\frac{\pi}{4}}\left(t^{6}-t^{4}\right) d t=\left(\frac{t^{7}}{7}-\frac{t^{5}}{5}\right)_{0}^{\frac{\pi}{4}}

\begin{aligned} &=\left(\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}\right)_{0}^{\frac{\pi}{4}}=\frac{1}{(\sqrt{2})^{7}} \times \frac{1}{7}-\frac{1}{(\sqrt{2})^{5}} \times \frac{1}{5}-\frac{1}{7}+\frac{1}{5} \\ & \end{aligned}

=\frac{1}{56 \sqrt{2}}-\frac{1}{20 \sqrt{2}}+\frac{2}{35}

\begin{aligned} &=\frac{1}{4 \sqrt{2}}\left(\frac{1}{14}-\frac{1}{5}\right)+\frac{2}{35} \\ & \end{aligned}

=\frac{2}{35}+\frac{1}{4 \sqrt{2}}\left(\frac{5-14}{70}\right) \\

=\frac{2}{35}-\frac{9}{280 \sqrt{2}}

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