If \;nx/\;x dx, then Iₙ =

Correct Answer is : −2n−1cos⁡(n−1)x−In−2-2/n-1 (n-1) x - Iₙ-2}n−1−2cos(n−1)x−In−2

Correct Answer is : −2n−1cos⁡(n−1)x−In−2-2/n-1 (n-1) x - Iₙ-2}n−1−2cos(n−1)x−In−2

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AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int \frac{\sin\;nx}{\cos\;x} dx,

I_n =

Solution:

Consider the integral.

I_n = \int \frac{\sin n x}{\cos x} dx

It is solved as,

I_n = \int \frac{\sin (n x - x + x)}{\cos x} dx

= \int \frac{\sin ((n-1)x + x)}{\cos x} dx

= \int \frac{\sin (n-1) x \cos x + \cos (n-1) x \sin x}{\cos x} dx

= \int \sin (n-1) dx + \int \frac{\cos (n-1) x \sin x}{\cos x} dx

Solve further,

I = \int \sin (n-1) dx

+ \frac{1}{2} \int \frac{2 \sin x \cos (n-1) x}{\cos x} dx

= - \frac{\cos (n-1)}{n-1} + \frac{1}{2} \int \frac{\sin n x + \sin (2-n) x}{\cos x} dx

= - \frac{\cos (n-1)}{n-1} + \frac{1}{2} \int \frac{\sin n x}{\cos x} dx

- \frac{1}{2} \int \frac{\sin (n-2) x}{\cos x} dx

= - \frac{\cos (n-1)}{n-1} + \frac{1}{2} I_n - \frac{1}{2} I_{n-2}

Solve further,

\left(1 - \frac{1}{2}\right) I_n = - \frac{\cos (n-1)}{n-1} - \frac{1}{2} I_{n-2}

I_n = \frac{-2}{n-1} \cos (n-1) x - I_{n-2}

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