The value of intlimits_0^{frac{pi }{2}} {sin | vedclass

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

Question

(D) Let $I = I_1 + I_2$,where $I_1 = \int_{0}^{\frac{\pi}{2}} \sin 8x \cot x \, dx$ and $I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln \left( \frac{

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(D) Let $I = I_1 + I_2$,where $I_1 = \int_{0}^{\frac{\pi}{2}} \sin 8x \cot x \, dx$ and $I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) dx$. For $I_2$,let $f(x) = \ln \left( \frac{1 - \sin x}{1 + \sin x} \right)$. Since $f(-x) = \ln \left( \frac{1 - \sin(-x)}{1 + \sin(-x)} \right) = \ln \left( \frac{1 + \sin x}{1 - \sin x} \right) = -\ln \left( \frac{1 - \sin x}{1 + \sin x} \right) = -f(x)$,the function is odd. Thus,$I_2 = 0$. Now,$I_1 = \int_{0}^{\frac{\pi}{2}} \frac{\sin 8x \cos x}{\sin x} \, dx$. Using the identity $\frac{\sin 8x}{\sin x} = 2(\cos 7x + \cos 5x + \cos 3x + \cos x)$,we have: $I_1 = 2 \int_{0}^{\frac{\pi}{2}} (\cos 7x + \cos 5x + \cos 3x + \cos x) \cos x \, dx$. Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $I_1 = \int_{0}^{\frac{\pi}{2}} (\cos 8x + \cos 6x + \cos 6x + \cos 4x + \cos 4x + \cos 2x + \cos 2x + 1) \, dx$. $I_1 = \int_{0}^{\frac{\pi}{2}} (\cos 8x + 2\cos 6x + 2\cos 4x + 2\cos 2x + 1) \, dx$. Integrating term by term: $I_1 = \left[ \frac{\sin 8x}{8} + \frac{2\sin 6x}{6} + \frac{2\sin 4x}{4} + \frac{2\sin 2x}{2} + x \right]_{0}^{\frac{\pi}{2}}$. Evaluating at the limits: $I_1 = (0 + 0 + 0 + 0 + \frac{\pi}{2}) - (0) = \frac{\pi}{2}$.

The value of $\int\limits_0^{\frac{\pi }{2}} {\sin 8x \cot x \, dx} + \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)dx}$ is equal to

Similar Questions

If $I = \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} dx$,then which of the following is true?

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

Let $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n x \, dx$. Then $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \dots$ are in:

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).

$\int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$ is equal to (where $\{.\}$ denotes the fractional part function and $n, r \in N$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module