Let In=∫0π/2 xncos⁡ x dx then I8+56I6 is equal to

In=∫0π/2 xncos⁡ xdx=xnsin⁡ x0π/2−n∫0π/2 xn−1sin⁡ xdx=π2n−n−xn−1cos⁡ x0π/2+(n−1)∫0π/2 xn−2cos⁡ xdx=π2n−n(n−1)In−2I8+8.7I6=π28

Let In=∫0π/2 xncos⁡ x dx then I8+56I6 is equal to

In=∫0π/2 xncos⁡ xdx=xnsin⁡ x0π/2−n∫0π/2 xn−1sin⁡ xdx=π2n−n−xn−1cos⁡ x0π/2+(n−1)∫0π/2 xn−2cos⁡ xdx=π2n−n(n−1)In−2I8+8.7I6=π28

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $I_{n} = \int_{0}^{π / 2} x^{n} \cos x d x$ then $I_{8} + 56 I_{6}$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${(\frac{π}{2})}^{6}$

${(\frac{π}{2})}^{8}$

${(\frac{π}{2})}^{6} - 1$

${(\frac{π}{2})}^{8} - 1$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I_{n} = \int_{0}^{π / 2} x^{n} \cos x d x \\ = {x^{n} \sin x|}_{0}^{π / 2} - n \int_{0}^{π / 2} x^{n - 1} \sin x d x \\ = {(\frac{π}{2})}^{n} - n [- {x^{n - 1} \cos x|}_{0}^{π / 2} + (n - 1) \int_{0}^{π / 2} x^{n - 2} \cos x d x] \\ = {(\frac{π}{2})}^{n} - n (n - 1) I_{n - 2} \\ I_{8} + 8.7 I_{6} = {(\frac{π}{2})}^{8} \end{array}$