Correct Answer is : 16log⁡(1−cos⁡x)+12log⁡(1+cos⁡x)1/6(1- x)+1/2(1+ x)61log(1−cosx)+21log(1+cosx) −23log⁡∣1+2cos⁡x∣+c-2/3≤ft|1+2 x|+c−32log∣1+2cosx∣+c

1/6(1- x)+1/2(1+ x) +2/3≤ft|1+2 x|+c

1/6(1- x)-1/2(1+ x) -2/3≤ft|1+2 x|+c

1/6(1- x)+1/2(1+ x) -2/3≤ft|1+2 x|+c

1/6[(1- x)(1+ x) ≤ft|1+2 x|]+c

Correct Answer is : 16log⁡(1−cos⁡x)+12log⁡(1+cos⁡x)1/6(1- x)+1/2(1+ x)61log(1−cosx)+21log(1+cosx) −23log⁡∣1+2cos⁡x∣+c-2/3≤ft|1+2 x|+c−32log∣1+2cosx∣+c

Test Index

AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

\int\frac{dx}{\sin x+\sin 2x}=

$\frac{1}{6}\log(1-\cos x)+\frac{1}{2}\log(1+\cos x)$ $+\frac{2}{3}\log\left|1+2\cos x\right|+c$
$\frac{1}{6}\log(1-\cos x)-\frac{1}{2}\log(1+\cos x)$ $-\frac{2}{3}\log\left|1+2\cos x\right|+c$
$\frac{1}{6}\log(1-\cos x)+\frac{1}{2}\log(1+\cos x)$ $-\frac{2}{3}\log\left|1+2\cos x\right|+c$
$\frac{1}{6}\log[(1-\cos x)(1+\cos x)$ $\left|1+2\cos x\right|]+c$

Validate

Solution:

Consider the expression,

I=\int\frac{dx}{\sin x+\sin 2x}

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

\int\frac{\sin x\,dx}{\sin^2 x(1+2\cos x)}

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

Let

\cos x = t

then,

-\sin x\,dx = dt

So,

I=-\int\frac{dt}{(1-t)(1+t)(1+2t)}

By partial fraction,

\frac{1}{(1-t)(1+t)(1+2t)} = \frac{A}{1-t}

+\frac{B}{1+t}+\frac{C}{1+2t}

1=A(1+t)(1+2t)+B(1-t)(1+2t)

+C(1+t)(1-t)

Then,

A=\frac{1}{6},\; B=-\frac{1}{2},\; C=\frac{4}{3}

So,

I=-\frac{1}{6}\int\frac{dt}{1-t}+\frac{1}{2}\int\frac{dt}{1+t}

-\frac{4}{3}\int\frac{dt}{1+2t}

\frac{1}{6}\log(1-\cos x)+\frac{1}{2}\log(1+\cos x)

-\frac{2}{3}\log\left|1+2\cos x\right|+c

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