The value of the integral int_{0}^{pi / 4} fr | vedclass

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

Question

(D) Let $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$, we can rewrite

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$, we can rewrite the denominator as $3 + 1 - (\sin x - \cos x)^2 = 4 - (\sin x - \cos x)^2$. Thus, $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{4 - (\sin x - \cos x)^2} dx$. Let $t = \sin x - \cos x$, then $dt = (\cos x + \sin x) dx$. When $x = 0$, $t = \sin 0 - \cos 0 = -1$. When $x = \pi / 4$, $t = \sin(\pi / 4) - \cos(\pi / 4) = 0$. Substituting these into the integral, we get $I = \int_{-1}^{0} \frac{dt}{4 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log |\frac{a+x}{a-x}|$, we have: $I = \frac{1}{2(2)} [\log |\frac{2+t}{2-t}|]_{-1}^{0} = \frac{1}{4} [\log |\frac{2+0}{2-0}| - \log |\frac{2-1}{2+1}|]$. $I = \frac{1}{4} [\log 1 - \log (1/3)] = \frac{1}{4} [0 - (-\log 3)] = \frac{1}{4} \log 3$.

The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

Similar Questions

$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $

$\int_{1/e}^e |\log x| \, dx = $

$\int_0^3 |x^2 - 3x + 2| dx = $

Prove that $\int_{1}^{3} \frac{dx}{x^{2}(x+1)} = \frac{2}{3} + \log \frac{2}{3}$.

The value of $\sum\limits_{r = 2}^{16} {\int\limits_r^{r + 1} {\frac{{dx}}{{\left( {2r - x} \right)\left( {2r + 2 - x} \right)}}} }$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module