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Question: If I<sub>n</sub> = \(\int_{0}^{\pi/2}x^{n}\)sinx dx then I<sub>7</sub> + 42 I<sub>5</sub> =...

If In = ∫0π/2xn\int_{0}^{\pi/2}x^{n}sinx dx then I7 + 42 I5 =

A

(π2)7\left( \frac{\pi}{2} \right)^{7}

B

(π2)6\left( \frac{\pi}{2} \right)^{6}

C

7(π2)6\left( \frac{\pi}{2} \right)^{6}

D

7(π2)7\left( \frac{\pi}{2} \right)^{7}

Answer

7(π2)6\left( \frac{\pi}{2} \right)^{6}

Explanation

Solution

In=∫0π/2xn\int_{0}^{\pi/2}x^{n}sinx dx = [xn(−cos⁡x)]0π/2\left\lbrack x^{n}( - \cos x) \right\rbrack_{0}^{\pi/2}–

∫0π/2nxn−1\int_{0}^{\pi/2}{nx^{n - 1}} (– cos x)dx

= n∫0π/2\int_{0}^{\pi/2}{}xn–1 cosx dx

= n [[xn−1(sin⁡x)]0π/2−(n−1)∫0π/2xn−2sin⁡xdx]\lbrack\lbrack x^{n - 1}(\sin x)\rbrack_{0}^{\pi/2} - (n - 1)\int_{0}^{\pi/2}x^{n - 2}\sin xdx\rbrack

In = n (π2)n−1\left( \frac{\pi}{2} \right)^{n - 1}– (n – 1)n In–2 put n = 7