If int frac{d x}{cos ^4 x+sin ^4 x}=frac{1}{s | vedclass

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

Question

(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+	an^4 x} d x

Vedclass · Accepted Answer

$\frac{	an x-\cot x}{\sqrt{2}}$

Vedclass · Answer

(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+	an^4 x} d x = \int \frac{\sec^2 x(1+	an^2 x)}{1+	an^4 x} d x$.Let $	an x = t$,then $\sec^2 x d x = d t$.$I = \int \frac{1+t^2}{1+t^4} d t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t$.$I = \int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2} d t$.Let $u = t-\frac{1}{t}$,then $d u = (1+\frac{1}{t^2}) d t$.$I = \int \frac{d u}{u^2+(\sqrt{2})^2} = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{u}{\sqrt{2}}) + C$.Substituting $u = t-\frac{1}{t} = 	an x - \cot x$,we get $I = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{	an x - \cot x}{\sqrt{2}}) + C$.Comparing this with the given form,$g(x) = \frac{	an x - \cot x}{\sqrt{2}}$.

If $\int \frac{d x}{\cos ^4 x+\sin ^4 x}=\frac{1}{\sqrt{2}} \tan ^{-1}[g(x)]+C$,then $g(x)$ equals

Similar Questions

$\int \frac{\sqrt{\cos 2 x}}{\sin x} d x=$

Let a function $h(x)$ be defined as $h(x) = 0$, for all $x \ne 0$. Also $\int_{-\infty}^{\infty} h(x) \cdot f(x) \, dx = f(0)$, for every function $f(x)$. Then the value of the definite integral $\int_{-\infty}^{\infty} h'(x) \cdot \sin x \, dx$ is

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

$\int \frac{1}{x^5 \sqrt[5]{x^5+1}} d x=$

For $x > 0$,if $\int (\log x)^5 dx$ is equal to $x[A(\log x)^5 + B(\log x)^4 + C(\log x)^3 + D(\log x)^2 + E(\log x) + F] + \text{constant}$,then $A + B + C + D + E + F$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module