If int frac{sin x}{sin 4x} dx = alpha log lef | vedclass

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

Question

(D) Given $I = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(D) Given $I = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x (1 - 2 \sin^2 x)}$.Multiply numerator and denominator by $\cos x$: $I = \frac{1}{4} \int \frac{\cos x dx}{\cos^2 x (1 - 2 \sin^2 x)} = \frac{1}{4} \int \frac{\cos x dx}{(1 - \sin^2 x)(1 - 2 \sin^2 x)}$.Let $u = \sin x$, then $du = \cos x dx$. $I = \frac{1}{4} \int \frac{du}{(1 - u^2)(1 - 2u^2)}$.Using partial fractions: $\frac{1}{(1 - u^2)(1 - 2u^2)} = \frac{A}{1 - u^2} + \frac{B}{1 - 2u^2}$.$1 = A(1 - 2u^2) + B(1 - u^2)$. For $u^2 = 1$, $A = -1$. For $u^2 = 1/2$, $B = 2$.$I = \frac{1}{4} \int \left( \frac{-1}{1 - u^2} + \frac{2}{1 - 2u^2} \right) du = -\frac{1}{4} \int \frac{du}{1 - u^2} + \frac{2}{4} \int \frac{du}{1 - (\sqrt{2}u)^2}$.$I = -\frac{1}{8} \log \left| \frac{1+u}{1-u} \right| + \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}u}{1-\sqrt{2}u} \right| + c$.Comparing with given form: $\alpha = -1/8$, $\beta = 1/(4\sqrt{2})$.$\beta^2 = 1/32$. Thus, $32(\alpha + \beta^2) = 32(-1/8 + 1/32) = 32(-4/32 + 1/32) = 32(-3/32) = -3$.

If $\int \frac{\sin x}{\sin 4x} dx = \alpha \log \left| \frac{1+\sin x}{1-\sin x} \right| + \beta \log \left| \frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x} \right| + c$, then the value of $32(\alpha + \beta^2) = $

Similar Questions

Evaluate the integral: $\int \frac{2x}{4 - 3x - x^2} dx$

$\int \frac{x+1}{x^{2}+5 x+6} d x=$

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

Integrate the function: $\frac{e^{x}}{(1+e^{x})(2+e^{x})}$

$\int_0^1 \frac{2 x+5}{x^2+3 x+2} \,d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module