If f(x) = int csc^5 x dx,then f(frac{pi}{4}) | vedclass

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

Question

(C) We have $f(x) = \int \csc^5 x \ dx$. Using the reduction formula for $\int \csc^n x \ dx$,we know that $\int \csc^n x \ dx = -\frac{\csc^{n-2} x \

Vedclass · Accepted Answer

$-\frac{1}{8} [7\sqrt{2} + 3\log(\sqrt{2} + 1)] + c$

Vedclass · Answer

(C) We have $f(x) = \int \csc^5 x \ dx$. Using the reduction formula for $\int \csc^n x \ dx$,we know that $\int \csc^n x \ dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} \int \csc^{n-2} x \ dx$. For $n=5$,we get $f(x) = -\frac{\csc^3 x \cot x}{4} + \frac{3}{4} \int \csc^3 x \ dx$. We know $\int \csc^3 x \ dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x| + C$. Substituting this back,$f(x) = -\frac{1}{4} \csc^3 x \cot x + \frac{3}{4} [-\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x|] + c$. $f(x) = -\frac{1}{4} \csc^3 x \cot x - \frac{3}{8} \csc x \cot x + \frac{3}{8} \log |\csc x - \cot x| + c$. At $x = \frac{\pi}{4}$,$\csc(\frac{\pi}{4}) = \sqrt{2}$ and $\cot(\frac{\pi}{4}) = 1$. $f(\frac{\pi}{4}) = -\frac{1}{4} (\sqrt{2})^3 (1) - \frac{3}{8} (\sqrt{2})(1) + \frac{3}{8} \log |\sqrt{2} - 1| + c$. $f(\frac{\pi}{4}) = -\frac{2\sqrt{2}}{4} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\sqrt{2} - 1| + c$. $f(\frac{\pi}{4}) = -\frac{4\sqrt{2}}{8} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}| + c$. $f(\frac{\pi}{4}) = -\frac{7\sqrt{2}}{8} + \frac{3}{8} \log |\frac{1}{\sqrt{2}+1}| + c = -\frac{7\sqrt{2}}{8} - \frac{3}{8} \log(\sqrt{2}+1) + c$. $f(\frac{\pi}{4}) = -\frac{1}{8} [7\sqrt{2} + 3 \log(\sqrt{2} + 1)] + c$.

If $f(x) = \int \csc^5 x \ dx$,then $f(\frac{\pi}{4}) = $

Similar Questions

If $\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{2\sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) - \frac{1}{2} \log g(x) + C$,then find the value of $g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)$.

If $\int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx\, = \,A(x)\,{(\sqrt {1 - {x^2}} )^m}\, + \,C,$ for a suitably chosen integer $m$ and a function $A(x),$ where $C$ is a constant of integration,then $(A(x))^m$ equals

If $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C, (a>0, b>0)$, then $\int \frac{d x}{b+a \cos x}=$

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$, then $f(x)$ is equal to

Find the integral of the function $\frac{1}{\cos (x-a) \cos (x-b)}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module