For n 0, int_0^{2pi } {frac{{x{{sin }^{2n}} | vedclass

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

Question

(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.Using the property $\int_0^a {f(x)dx = \int_0^a {f(a

Vedclass · Accepted Answer

${\pi ^2}$

Vedclass · Answer

(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.Using the property $\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} }$,we have:$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}(2\pi - x)}}{{{{\sin }^{2n}}(2\pi - x) + {{\cos }^{2n}}(2\pi - x)}}dx} $$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $Adding the two expressions for $I$:$2I = \int_0^{2\pi } {\frac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$I = \pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $Using the property $\int_0^{2a} {f(x)dx = 2\int_0^a {f(x)dx} }$ if $f(2a-x) = f(x)$:$I = 2\pi \int_0^{\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $i$)Similarly,$I = 4\pi \int_0^{\pi /2} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $ii$)Adding $(i)$ and $(ii)$:$2I = 4\pi \int_0^{\pi /2} {1 dx} = 4\pi \left( \frac{\pi }{2} \right) = 2{\pi ^2}$Therefore,$I = {\pi ^2}$.

For $n > 0,$ $\int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}\,dx = } $

Similar Questions

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

$\int_{0}^{\pi} \cos^3 x \, dx = $

$\int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$ is equal to

Evaluate the definite integral: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}$.

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module