int frac{1}{cos x} left[ frac{1}{sin x} - fra | vedclass

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

Question

(D) Let $I = \int \frac{1}{\cos x} \left[ \frac{\sin x + 3 \cos x - \sin x}{\sin x(\sin x + 3 \cos x)} \right] dx$ $= \int \frac{1}{\cos x} \left[ \fr

Vedclass · Accepted Answer

$\log \left| \frac{\sin x}{\sin x + 3 \cos x} \right| + c$

Vedclass · Answer

(D) Let $I = \int \frac{1}{\cos x} \left[ \frac{\sin x + 3 \cos x - \sin x}{\sin x(\sin x + 3 \cos x)} ight] dx$ $= \int \frac{1}{\cos x} \left[ \frac{3 \cos x}{\sin x(\sin x + 3 \cos x)} ight] dx$ $= \int \frac{3}{\sin x(\sin x + 3 \cos x)} dx$ Divide numerator and denominator by $\cos^2 x$: $= \int \frac{3 \sec^2 x}{	an x(	an x + 3)} dx$ Let $u = 	an x$,then $du = \sec^2 x dx$. $I = \int \frac{3}{u(u + 3)} du$ Using partial fractions: $\frac{3}{u(u + 3)} = \frac{A}{u} + \frac{B}{u + 3} \implies 3 = A(u + 3) + Bu$. For $u = 0$,$A = 1$. For $u = -3$,$B = -1$. $I = \int \left( \frac{1}{u} - \frac{1}{u + 3} ight) du = \log |u| - \log |u + 3| + c$ $= \log \left| \frac{u}{u + 3} ight| + c = \log \left| \frac{	an x}{	an x + 3} ight| + c$ $= \log \left| \frac{\sin x / \cos x}{(\sin x + 3 \cos x) / \cos x} ight| + c = \log \left| \frac{\sin x}{\sin x + 3 \cos x} ight| + c$.

$\int \frac{1}{\cos x} \left[ \frac{1}{\sin x} - \frac{1}{\sin x + 3 \cos x} \right] dx =$

Similar Questions

$\int \frac{dx}{1 + x + x^2 + x^3} = $

$ \int \frac{x^2+1}{x(x^2-1)} dx $

Find $\int \frac{dx}{(x+1)(x+2)}$

Let $f(x) = \int \frac{16x + 24}{x^2 + 2x - 15} dx$. If $f(4) = 14 \log_e(3)$ and $f(7) = \log_e(2^\alpha \cdot 3^\beta)$, where $\alpha, \beta \in N$, then $\alpha + \beta$ is equal to:

$\int \frac{3x+4}{x^3-2x-4} dx = \log f(x) + C \Rightarrow f(3) = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module