If int frac{1}{(sin x + 4)(sin x - 1)} dx = A | vedclass

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

Question

(D) We have the integral $I = \int \frac{1}{(\sin x + 4)(\sin x - 1)} dx$.Using partial fractions,$\frac{1}{(\sin x + 4)(\sin x - 1)} = \frac{1}{5} \l

Vedclass · Accepted Answer

$A = \frac{2}{5}, B = -\frac{2}{5\sqrt{15}}, f(x) = \frac{4	an \frac{x}{2} + 1}{\sqrt{15}}$

Vedclass · Answer

(D) We have the integral $I = \int \frac{1}{(\sin x + 4)(\sin x - 1)} dx$.Using partial fractions,$\frac{1}{(\sin x + 4)(\sin x - 1)} = \frac{1}{5} \left( \frac{1}{\sin x - 1} - \frac{1}{\sin x + 4} ight)$.Substituting $\sin x = \frac{2t}{1+t^2}$ where $t = 	an \frac{x}{2}$ and $dx = \frac{2dt}{1+t^2}$,we get:$I = \frac{1}{5} \int \frac{1}{\frac{2t}{1+t^2} - 1} \frac{2dt}{1+t^2} - \frac{1}{5} \int \frac{1}{\frac{2t}{1+t^2} + 4} \frac{2dt}{1+t^2}$$I = \frac{1}{5} \int \frac{2dt}{2t - 1 - t^2} - \frac{1}{5} \int \frac{2dt}{2t + 4 + 4t^2}$$I = -\frac{2}{5} \int \frac{dt}{(t-1)^2} - \frac{1}{10} \int \frac{dt}{t^2 + \frac{1}{2}t + 1}$$I = \frac{2}{5(t-1)} - \frac{1}{10} \int \frac{dt}{(t + \frac{1}{4})^2 + (\frac{\sqrt{15}}{4})^2}$$I = \frac{2}{5(	an \frac{x}{2} - 1)} - \frac{1}{10} \cdot \frac{4}{\sqrt{15}} 	an^{-1} \left( \frac{t + 1/4}{\sqrt{15}/4} ight) + C$$I = \frac{2}{5(	an \frac{x}{2} - 1)} - \frac{2}{5\sqrt{15}} 	an^{-1} \left( \frac{4	an \frac{x}{2} + 1}{\sqrt{15}} ight) + C$.Comparing this with the given form,we get $A = \frac{2}{5}, B = -\frac{2}{5\sqrt{15}}, f(x) = \frac{4	an \frac{x}{2} + 1}{\sqrt{15}}$.

If $\int \frac{1}{(\sin x + 4)(\sin x - 1)} dx = A \frac{1}{\tan \frac{x}{2} - 1} + B \tan^{-1}(f(x)) + C$,then

Similar Questions

$\int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx = A \sqrt{7-6x-x^2} + B \sin^{-1}\left(\frac{x+3}{4}\right) + c$ (where $c$ is a constant of integration),then the value of $A+B$ is

If $\int(3 x+2) \sqrt{2 x^2+3 x+4} d x=f(x) \sqrt{2 x^2+3 x+4}+A \sinh ^{-1}\left(\frac{4 x+3}{\sqrt{23}}\right)+C$, then the ordered pair $(f(1), A)=$

$\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=$

Evaluate the integral: $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx$

Evaluate the integral $\int \frac{x^2 + \cos^2 x}{1 + x^2} \operatorname{cosec}^2 x \, dx$,where $c$ is the constant of integration.

Vedclass Test Series

Exam Paper Generator

Online Exam Module