If In=∫sin2⁡nxsin2⁡xdx then In+1−2In+In−1=

In=∫sin2⁡nxsin2⁡xdxIn+1−In=∫sin2⁡(n+1)x−sin2⁡nxsin2xdx=∫sin⁡(2n+1)xsin⁡xsin2⁡xdx=∫sin⁡(2n+1)xsin⁡xdxnow,In−In−1=∫sin⁡(2(n−1)+1)xsin⁡x=∫sin⁡(2n−1)xsin⁡xdx In+1−In−In−In−1=∫sin⁡(2n+1)x−sin⁡(2n−1)xsin⁡xdxIn+1−2In+In-1=∫2cos⁡2nx sin⁡xsin⁡xdx=2sin⁡2nx2n+c=1nSin2nx+C

If In=∫sin2⁡nxsin2⁡xdx then In+1−2In+In−1=

In=∫sin2⁡nxsin2⁡xdxIn+1−In=∫sin2⁡(n+1)x−sin2⁡nxsin2xdx=∫sin⁡(2n+1)xsin⁡xsin2⁡xdx=∫sin⁡(2n+1)xsin⁡xdxnow,In−In−1=∫sin⁡(2(n−1)+1)xsin⁡x=∫sin⁡(2n−1)xsin⁡xdx In+1−In−In−In−1=∫sin⁡(2n+1)x−sin⁡(2n−1)xsin⁡xdxIn+1−2In+In-1=∫2cos⁡2nx sin⁡xsin⁡xdx=2sin⁡2nx2n+c=1nSin2nx+C

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

$If I_{n} = \int \frac{\sin^{2} n x}{\sin^{2} x} d x then I_{n + 1} - 2 I_{n} + I_{n - 1} =$

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

$- \frac{1}{n} \cos 2 n x + c$

$- \frac{1}{n} \sin 2 n x + c$

$\frac{1}{n} \sin 2 n x + c$

$\frac{1}{n} \cos 2 n x + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I_{n} = \int \frac{\sin^{2} n x}{\sin^{2} x} d x$
$I_{n + 1} - I_{n} = \int \frac{\sin^{2} (n + 1) x - \sin^{2} n x}{\sin^{2} x} d x$
$\begin{aligned} = \int \frac{\sin (2 n + 1) x \sin x}{\sin^{2} x} d x \\ = \int \frac{\sin (2 n + 1) x}{\sin x} d x \end{aligned}$
now,
$\begin{aligned} I_{n} - I_{n - 1} = \int \frac{\sin (2 (n - 1) + 1) x}{\sin x} \\ = \int \frac{\sin (2 n - 1) x}{\sin x} d x \end{aligned}$
$\begin{array}{l} \begin{aligned} (I_{n + 1} - I_{n}) - (I_{n} - I_{n - 1}) = \int \frac{\sin (2 n + 1) x - \sin (2 n - 1) x}{\sin x} d x \end{aligned} \\ I_{n + 1} - 2 I_{n} + I_{n - 1} = \int \frac{2 \cos 2 n x \sin x}{\sin x} d x \\ = 2 \frac{\sin 2 n x}{2 n} + c \\ = \frac{1}{n} S i n 2 n x + C \end{array}$