If int operatorname{cosec}^5 x , dx = alpha c | vedclass

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

Question

(B) Let $I = \int \operatorname{cosec}^5 x \, dx$. Using integration by parts with $u = \operatorname{cosec}^3 x$ and $dv = \operatorname{cosec}^2 x \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $I = \int \operatorname{cosec}^5 x \, dx$. Using integration by parts with $u = \operatorname{cosec}^3 x$ and $dv = \operatorname{cosec}^2 x \, dx$:$I = \operatorname{cosec}^3 x(-\cot x) - \int (3 \operatorname{cosec}^2 x(-\operatorname{cosec} x \cot x))(-\cot x) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x \cot^2 x \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x (\operatorname{cosec}^2 x - 1) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3I + 3 \int \operatorname{cosec}^3 x \, dx$$4I = -\operatorname{cosec}^3 x \cot x + 3 \int \operatorname{cosec}^3 x \, dx$We know $\int \operatorname{cosec}^3 x \, dx = -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}|$.Substituting this into the equation for $4I$:$4I = -\operatorname{cosec}^3 x \cot x + 3 \left( -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}| ight)$$4I = -\operatorname{cosec}^3 x \cot x - \frac{3}{2} \operatorname{cosec} x \cot x + \frac{3}{2} \ln |	an \frac{x}{2}|$$I = -\frac{1}{4} \cot x \operatorname{cosec} x (\operatorname{cosec}^2 x + \frac{3}{2}) + \frac{3}{8} \ln |	an \frac{x}{2}| + C$Comparing with the given form,$\alpha = -\frac{1}{4}$ and $\beta = \frac{3}{8}$.Thus,$8(\alpha + \beta) = 8(-\frac{1}{4} + \frac{3}{8}) = 8(\frac{-2+3}{8}) = 1$.

If $\int \operatorname{cosec}^5 x \, dx = \alpha \cot x \operatorname{cosec} x \left(\operatorname{cosec}^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$,where $\alpha, \beta \in R$ and $C$ is the constant of integration,then the value of $8(\alpha + \beta)$ is equal to:

Similar Questions

Evaluate the integral $\int \frac{x^2 + \cos^2 x}{1 + x^2} \operatorname{cosec}^2 x \, dx$,where $c$ is the constant of integration.

If $\int \frac{1}{(\sin x + 4)(\sin x - 1)} dx = A \frac{1}{\tan \frac{x}{2} - 1} + B \tan^{-1}(f(x)) + C$,then

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

If $\int(x^6+x^4+x^2) \sqrt{2x^4+3x^2+6} dx = f(x) + c$,then $f(3) =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module