If Iₙ = ∫ nx/ x dx for n = 1 , 2 , 3 , ... , then I₆ =

Correct Answer is : 252/552 sin 5x - 85sin⁡38/5 ³58sin3 x + 4 sin x + x

3/5 sin x + 8/3 ^5 x - sin x + c

2/5 sin 5x - 5/3 ³ x - 2 sin x + c

2/3 sin 5x - 8/3 ^5 x + 4 sin x + c

2/5 sin 5x - 8/5 ³ x + 4 sin x + x

Correct Answer is : 252/552 sin 5x - 85sin⁡38/5 ³58sin3 x + 4 sin x + x

Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

I_n

\frac{\sin nx}{\sin x}

I_6

Solution:

Consider the given expression

I_{n} = \int\; \frac{\sin n x}{\sin x} d x \;\;\;\;\; \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot (I)

And

I_{n-2} = \int\; \frac{\sin (n-2) x}{\sin x} d x \;\;\;\;\; \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot (II)

Subtract equation (I) and (II),

I_{n} - I_{n-2} \;\; = \int\; \frac{\sin n x - \sin (n-2) x}{\sin x} d x

\;\; = \int\; \frac{2 \sin (n-1) x}{\sin x} d x

\;\; = 2 \cos (n-1) x \, d x

\;\; = \frac{2 \sin (n-1) x}{n-1}

This implies,

I_{6} - I_{4} = \frac{2 \sin 5x}{5} \text{ and } I_{4} - I_{2} = \frac{2 \sin 3x}{3}

Now,

I_{2} \;\; = \int\; \frac{\sin 2x}{\sin x} d x

\;\; = \int\; \frac{2 \sin x \cos x}{\sin x} d x

\;\; = 2 \int \cos x \, d x

\;\; = 2 \sin x + c

And,

I_{6} \;\; = I_{4} + 2 \frac{\sin 5x}{5}

\;\; = I_{2} + 2 \frac{\sin 3x}{3} + 2 \frac{\sin 5x}{5}

\;\; = 2 \frac{\sin 5x}{5} + 2 \frac{\sin 3x}{3} + 2 \sin x + c

\;\; = \frac{2 \sin 5x}{5} + \frac{2}{3}(3 \sin x - 4 \sin^3 x) + 2 \sin x + c

I_{6} = \frac{2}{5} \sin 5x - \frac{8}{3} \sin^3 x + 4 \sin x + c

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