NDA 2015
Exam NDA
Year 2015
Subject Mathematics
Chapter Definite Integration NDA 2015 · Mathematics · Definite Integration
∫_0^ 2 π ^5((x)/(4)) d x is equal to
∫ 0 2 π sin 5 ( x 4 ) d x \int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x ∫ 0 2 π sin 5 ( 4 x ) d x is equal to
Options A 8/15 B 16/15 C 32/15 D 0
Step-by-step solution
∫ 0 2 π sin 5 ( x 4 ) d x = ∫ 0 2 π ( 1 − cos 2 x 4 ) ( 1 − cos 2 x 4 ) sin x 4 ⋅ d x \int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x=\int_0^{2 \pi}\left(1-\cos ^2 \frac{x}{4}\right)\left(1-\cos ^2 \frac{x}{4}\right) \sin \frac{x}{4} \cdot d x ∫ 0 2 π sin 5 ( 4 x ) d x = ∫ 0 2 π ( 1 − cos 2 4 x ) ( 1 − cos 2 4 x ) sin 4 x ⋅ d x
Put
cos ( x 4 ) = t \cos \left(\frac{x}{4}\right)=t cos ( 4 x ) = t ⇒ − sin ( x 4 ) ⋅ d x 4 = d t ⇒ sin ( x 4 ) d x = − 4 d t ⇒ ∫ 0 2 π sin 5 ( x 4 ) d x = − 4 ∫ ( 1 − t 2 ) ( 1 − t 2 ) d t = − 4 [ ( 1 + t 4 − 2 t 2 ) d t = − 4 [ t + t 5 5 − 2 t 3 3 ] = − 4 [ cos ( x 4 ) + cos 5 ( x 4 ) 5 − 2 cos 3 ( x 4 ) 3 ] 0 2 π \begin{aligned}
& \Rightarrow-\sin \left(\frac{x}{4}\right) \cdot \frac{d x}{4}=d t \\
& \Rightarrow \sin \left(\frac{x}{4}\right) d x=-4 d t \\
& \Rightarrow \int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x=-4 \int\left(1-t^2\right)\left(1-t^2\right) d t \\
& =-4\left[\left(1+t^4-2 t^2\right) d t\right. \\
& =-4\left[t+\frac{t^5}{5}-\frac{2 t^3}{3}\right] \\
& =-4\left[\cos \left(\frac{x}{4}\right)+\frac{\cos ^5\left(\frac{x}{4}\right)}{5}-\frac{2 \cos ^3\left(\frac{x}{4}\right)}{3}\right]_0^{2 \pi}
\end{aligned} ⇒ − sin ( 4 x ) ⋅ 4 d x = d t ⇒ sin ( 4 x ) d x = − 4 d t ⇒ ∫ 0 2 π sin 5 ( 4 x ) d x = − 4 ∫ ( 1 − t 2 ) ( 1 − t 2 ) d t = − 4 [ ( 1 + t 4 − 2 t 2 ) d t = − 4 [ t + 5 t 5 − 3 2 t 3 ] = − 4 [ cos ( 4 x ) + 5 cos 5 ( 4 x ) − 3 2 cos 3 ( 4 x ) ] 0 2 π = − 4 [ ( 0 + 0 − 0 ) − ( 1 + 1 5 − 2 3 ) ] = 32 15 =-4\left[(0+0-0)-\left(1+\frac{1}{5}-\frac{2}{3}\right)\right]=\frac{32}{15} = − 4 [ ( 0 + 0 − 0 ) − ( 1 + 5 1 − 3 2 ) ] = 15 32
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