If In-sin nxcos xdx- then In-In-2 is-where n-n-2 and c is constant of integration

Given: I_n=∫sin nxcos xdx Let us consider I_n+I_n 2 ⇒ I_n+I_n 2=∫sin nx+sin (n 2)xcos x dx ⇒ I_n+I_n 2=∫2 sin (n 1)x · nbsp; cos xcos x dx ⇒ I_n+I_n 2=∫ nbs ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If In=∫sin nx...

Question

If $I_{n} = \int \frac{sin n x}{cos x} d x$ , then $I_{n} + I_{n - 2}$ is:
(where $n \in N, n \geq 2$ and $c$ 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} cos (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 A $\frac{- 2}{n - 1} cos (n - 1) x + c$
Given: $I_{n} = \int \frac{sin n x}{cos x} d x$
Let us consider $I_{n} + I_{n - 2}$
$\Rightarrow I_{n} + I_{n - 2} = \int \frac{sin n x + sin (n - 2) x}{cos x} d x$
$\Rightarrow I_{n} + I_{n - 2} = \int \frac{2 sin (n - 1) x \cdot cos x}{cos x} d x$
$\Rightarrow I_{n} + I_{n - 2} = \int 2 sin (n - 1) x d x$
$\Rightarrow I_{n} + I_{n - 2} = \frac{- 2}{n - 1} cos (n - 1) x + c$

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Sum of Coefficients of All Terms

Standard XII Mathematics

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If In=∫sinnxcosxdx, then In+In−2 is: (where n∈N,n≥2 and c is constant of integration)

The correct option is A −2n−1cos(n−1)x+cGiven: In=∫sinnxcosxdx Let us consider In+In−2 ⇒In+In−2=∫sinnx+sin(n−2)xcosx dx ⇒In+In−2=∫2sin(n−1)x⋅cosxcosx dx ⇒In+In−2=∫2sin(n−1)x dx ⇒In+In−2=−2n−1cos(n−1)x+c

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