If In-sin nxsin x dx where n-1 and n - then In-In-2-where C is integration constant

Given: nbsp;I_n=∫sin nxsin x dx ⇒ I_n I_n 2= ∫sin nxsin x dx ∫sin (n 2)xsin x dx =∫sin nx sin (n 2)xsin x dx =∫2sin x. cos (n 1)x sin x dx =2∫cos (n 1)x dx = ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

If In=∫sin nx...

Question

If $I_{n} = \int \frac{sin n x}{sin x} d x$ where $n > 1$ and $n \in N,$ then $I_{n} - I_{n - 2} =$
(where $C$ is integration constant)

\frac{2}{n - 1} cos (n - 1) x + C

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\frac{2}{n - 1} sin (n - 1) x + C

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- \frac{2}{n - 1} sin (n - 1) x + C

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- \frac{2}{n - 1} cos (n - 1) x + C

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Solution

The correct option is B $\frac{2}{n - 1} sin (n - 1) x + C$
Given: $I_{n} = \int \frac{sin n x}{sin x} d x$
$\Rightarrow I_{n} - I_{n - 2} = \int \frac{sin n x}{sin x} d x - \int \frac{sin (n - 2) x}{sin x} d x$
$= \int \frac{sin n x - sin (n - 2) x}{sin x} d x$
$= \int \frac{2 sin x . cos (n - 1) x}{sin x} d x$
$= 2 \int cos (n - 1) x d x$
$= \frac{2}{n - 1} sin (n - 1) x + C$

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Inverse of a Matrix

Standard XII Mathematics

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If In=∫sinnxsinxdx where n>1 and n∈N, then In−In−2= (where C is integration constant)

The correct option is B 2n−1sin(n−1)x+CGiven: In=∫sinnxsinxdx ⇒In−In−2=∫sinnxsinxdx−∫sin(n−2)xsinxdx =∫sinnx−sin(n−2)xsinxdx =∫2sinx.cos(n−1)xsinx dx =2∫cos(n−1)x dx =2n−1sin(n−1)x+C

If $I_{n} = \int \frac{sin n x}{sin x} d x$ where $n > 1$ and $n \in N,$ then $I_{n} - I_{n - 2} =$
(where $C$ is integration constant)