int_0^{frac{pi}{4}} frac{sin x+cos x}{3+sin 2 | vedclass

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

Question

(D) Let $I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x = 0$,$

Vedclass · Accepted Answer

$\frac{1}{4} \log 3$

Vedclass · Answer

(D) Let $I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$. Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$. Substituting these into the integral: $I = \int_{-1}^0 \frac{dt}{3 + (1 - t^2)} = \int_{-1}^0 \frac{dt}{4 - t^2} = \int_{-1}^0 \frac{dt}{2^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$: $I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^0 = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$. $I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

Similar Questions

The value of the integral $\int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{x}\right)\left(\sin ^{6} x+\cos ^{6} x\right)}$ is equal to

The value of $\int_{ - a}^a {\frac{1}{{x + {x^3}}}} dx$ is

By using the properties of definite integrals,evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \, dx$.

If $f(a+b+1-x)=f(x)$ for all $x,$ where $a$ and $b$ are fixed positive real numbers,then $\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx$ is equal to:

$\int_{-1}^1 x|x| \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module