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

If $\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = A\left(\frac{\alpha x-1}{\beta x+3}\right)^B + C,$ where $C$ is the constant of integration,then the value of $\alpha + \beta + 20AB$ is...........

For $x \geq 0$, $\int \sqrt{x^2+2x} \, dx$ is equal to

$\text{If } \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{e}\left(4 e^{x}+7 e^{-x}\right)\right)+C$ where $C$ is a constant of integration,then $u+v$ is equal to .... .

If $\int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}} = A \cdot \tan^{-1} \sqrt{f(x)} + \text{constant}$, then the ordered pair $(A, f(-1)) =$

For $0 < x < 1$,evaluate the integral $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module