If In-cos nxcos x dx where n-1 and n - then In-In-2-where c is constant of integration

Given: nbsp;I_n=∫cos nxcos x dx ⇒ I_n+I_n 2=∫cos nx+cos(n 2)xcos x dx ⇒ I_n+I_n 2=∫2cos(n 1)xcos xcos x dx ⇒ I_n+I_n 2 =∫ 2cos(n 1)x dx ⇒ I_n+I_n 2= 2n 1sin ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

If In=∫cos nx...

Question

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

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

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\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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Solution

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

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Inverse of a Matrix

Standard XII Mathematics

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If In=∫cosnxcosxdx where n>1 and n∈N, then In+In−2= (where c is constant of integration)

The correct option is C 2n−1sin(n−1)x+cGiven: In=∫cosnxcosxdx ⇒In+In−2=∫cosnx+cos(n−2)xcosxdx⇒In+In−2=∫2cos(n−1)xcosxcosx dx⇒In+In−2=∫2cos(n−1)x dx⇒In+In−2=2n−1sin(n−1)x+c

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