If In=∫(sin⁡x+cos⁡x)ndx, then nIn=f(x)+2(n−1)In−2 where ∫f(x)dx=

In =∫(sin⁡x+cos⁡x)n−1(sin⁡x+cos⁡x)dx=(s+c)n−1(s−c)+∫(n−1)(s+c)n−2(c−s)(c−s)dx=(s+c)n−1(s−c)+(n−1)∫(s+c)n−2−(s+c)2+2dx=(s+c)n−1(s−c)+2(n−1)In−2−(n−1)In⇒n⋅In=(s+c)n−1(s−c)+2(n−1)In−2

If In=∫(sin⁡x+cos⁡x)ndx, then nIn=f(x)+2(n−1)In−2 where ∫f(x)dx=

In =∫(sin⁡x+cos⁡x)n−1(sin⁡x+cos⁡x)dx=(s+c)n−1(s−c)+∫(n−1)(s+c)n−2(c−s)(c−s)dx=(s+c)n−1(s−c)+(n−1)∫(s+c)n−2−(s+c)2+2dx=(s+c)n−1(s−c)+2(n−1)In−2−(n−1)In⇒n⋅In=(s+c)n−1(s−c)+2(n−1)In−2

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

If $I_{n} = \int (\sin x + \cos x)^{n} dx, then {nI}_{n} = f (x) + 2 (n - 1) I_{n - 2} where \int f (x) dx =$

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

$\frac{- (\sin x + \cos x)^{n}}{n} + c$

$\frac{(\sin x - \cos x)^{n + 1}}{n - 1} + c$

$\frac{(\cos x - \sin x)^{n}}{n} + c$

$\frac{- (\sin x + \cos x)^{n + 1}}{n + 1} + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} In = \int (\sin x + \cos x)^{n - 1} (\sin x + \cos x) dx \\ = (s + c)^{n - 1} (s - c) + \int (n - 1) (s + c)^{n - 2} (c - s) (c - s) dx \\ = (s + c)^{n - 1} (s - c) + (n - 1) \int (s + c)^{n - 2} [- (s + c)^{2} + 2] dx \end{array}$
$= (s + c)^{n - 1} (s - c) + 2 (n - 1) I_{n - 2} - (n - 1) I_{n} \Rightarrow n \cdot I_{n} = (s + c)^{n - 1} (s - c) + 2 (n - 1) I_{n - 2}$