Evaluate the integral: int frac{dx}{sin x + s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we get $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.Multiply numerato

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we get $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.Multiply numerator and denominator by $\sin x$: $I = \int \frac{\sin x dx}{\sin^2 x(1 + 2 \cos x)} = \int \frac{\sin x dx}{(1 - \cos^2 x)(1 + 2 \cos x)} = \int \frac{\sin x dx}{(1 - \cos x)(1 + \cos x)(1 + 2 \cos x)}$.Let $\cos x = t$,then $-\sin x dx = dt$,so $I = -\int \frac{dt}{(1-t)(1+t)(1+2t)}$.Using partial fractions: $\frac{1}{(1-t)(1+t)(1+2t)} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{1+2t}$.Solving for constants: $A = \frac{1}{6}$,$B = \frac{1}{2}$,$C = -\frac{4}{3}$.Thus,$I = -\int (\frac{1/6}{1-t} + \frac{1/2}{1+t} - \frac{4/3}{1+2t}) dt = \frac{1}{6} \log |1-t| - \frac{1}{2} \log |1+t| + \frac{4}{3} \cdot \frac{1}{2} \log |1+2t| + c$.Substituting $t = \cos x$: $I = \frac{1}{6} \log |1-\cos x| - \frac{1}{2} \log |1+\cos x| + \frac{2}{3} \log |1+2 \cos x| + c$.

Evaluate the integral: $\int \frac{dx}{\sin x + \sin 2x}$

Similar Questions

Integrate the rational function:
$\frac{x^{3}+x+1}{x^{2}-1}$

$\int \frac{2 x^3-4 x^2-x-3}{x^2-2 x-3} d x=$

$\int \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)} d x=$

$\int \frac{dx}{x(x^2+1)^3} = ?$

$\int \frac{d x}{2 e^{2 x}+3 e^x+1}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module