If int cos x cdot cos 2 x cdot cos 5 x , dx = | vedclass

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

Question

(B) Given,$\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx$ $= \frac{1}{2} \int (2 \cos 5 x \cos x) \cos 2 x \, dx$ $= \frac{1}{2} \int (\cos 6 x + \c

Vedclass · Accepted Answer

$\frac{1}{A} + \frac{1}{D}$

Vedclass · Answer

(B) Given,$\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx$ $= \frac{1}{2} \int (2 \cos 5 x \cos x) \cos 2 x \, dx$ $= \frac{1}{2} \int (\cos 6 x + \cos 4 x) \cos 2 x \, dx$ $= \frac{1}{4} \int (2 \cos 6 x \cos 2 x + 2 \cos 4 x \cos 2 x) \, dx$ $= \frac{1}{4} \int (\cos 8 x + \cos 4 x + \cos 6 x + \cos 2 x) \, dx$ $= \frac{1}{4} \left[ \frac{\sin 8 x}{8} + \frac{\sin 4 x}{4} + \frac{\sin 6 x}{6} + \frac{\sin 2 x}{2} \right] + k$ $= \frac{\sin 2 x}{8} + \frac{\sin 4 x}{16} + \frac{\sin 6 x}{24} + \frac{\sin 8 x}{32} + k$ Comparing with $A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$,we get: $A = \frac{1}{8}, B = \frac{1}{16}, C = \frac{1}{24}, D = \frac{1}{32}$ Then,$\frac{1}{B} + \frac{1}{C} = 16 + 24 = 40$ And $\frac{1}{A} + \frac{1}{D} = 8 + 32 = 40$ Therefore,$\frac{1}{B} + \frac{1}{C} = \frac{1}{A} + \frac{1}{D}$.

If $\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx = A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$ (where $k$ is the arbitrary constant of integration),then $\frac{1}{B} + \frac{1}{C} = $

Similar Questions

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ is equal to

$\int \frac{dx}{\cos 2x + \sin^2 x} = $

Find an anti-derivative (or integral) of the function $\sin 2x$ by the method of inspection.

Find the following integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

$\int \left( \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} \right) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module