int frac{sin x+sin ^3 x}{cos 2 x} ,d x=A cos | vedclass

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

Question

(A) Let $I = \int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x$. $= \int \frac{\sin x(1+\sin ^2 x)}{\cos 2 x} \,d x$. $= \int \frac{\sin x(1+1-\cos ^2 x)}{

Vedclass · Accepted Answer

$A=\frac{1}{2}, B=\frac{-3}{4 \sqrt{2}}, f(x)=\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x$. $= \int \frac{\sin x(1+\sin ^2 x)}{\cos 2 x} \,d x$. $= \int \frac{\sin x(1+1-\cos ^2 x)}{2 \cos ^2 x-1} \,d x$. $= \int \frac{\sin x(2-\cos ^2 x)}{2 \cos ^2 x-1} \,d x$. Substitute $\cos x = t$,so $-\sin x \,d x = dt$ or $\sin x \,d x = -dt$. $I = -\int \frac{2-t^2}{2 t^2-1} dt = \int \frac{t^2-2}{2 t^2-1} dt$. $= \frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} dt = \frac{1}{2} \int \left(1 - \frac{3}{2 t^2-1}\right) dt$. $= \frac{1}{2} \int dt - \frac{3}{2} \int \frac{dt}{(\sqrt{2} t)^2-1^2}$. Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log |\frac{x-a}{x+a}|$,we get: $= \frac{1}{2} t - \frac{3}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \log |\frac{\sqrt{2} t-1}{\sqrt{2} t+1}| + c$. $= \frac{1}{2} \cos x - \frac{3}{4 \sqrt{2}} \log |\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}| + c$. Comparing with $A \cos x + B \log |f(x)| + c$,we have $A = \frac{1}{2}$,$B = \frac{-3}{4 \sqrt{2}}$,and $f(x) = \frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}$.

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=A \cos x+B \log |f(x)|+c$ (where $c$ is a constant of integration). Then the values of $A, B$ and $f(x)$ are:

Similar Questions

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

$\int \frac{2x^2-1+x^2\sqrt{x^2+4}}{x^2(x^2+4)} dx =$

If $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ then $\frac{\lambda}{\mu}=$

$\begin{aligned} & \text{If } 5(f(x))^2 = x f(x) + 30 \text{ and } \\ & \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx \\ & = \frac{A}{B x^3 + D f(x)} + C, \text{ then } A + B + D = \end{aligned}$

If $\int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{x} + e^{-x})} dx = g(x) e^{(e^{x} + e^{-x})} + c$,where $c$ is a constant of integration,then $g(0)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module