Advertisements
Advertisements
Question
\[\int \sin^2\text{ b x dx}\]
Sum
Advertisements
Solution
\[\int \sin^2\text{ bx dx}\]
\[ = \int\left[ \frac{1 - \cos 2bx}{2} \right]dx \left[ \therefore \sin^2 x = \frac{1 - \cos 2x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos 2bx \right)dx\]
\[ = \frac{1}{2}\left[ x - \frac{\sin 2bx}{2b} \right] + C\]
\[ = \frac{x}{2} - \frac{\sin 2bx}{4b} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
` ∫ tan x sec^4 x dx `
\[\int \sin^5 x \text{ dx }\]
\[\int \cos^7 x \text{ dx } \]
\[\int\frac{1}{1 + x - x^2} \text{ dx }\]
\[\int\frac{e^x}{1 + e^{2x}} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int x \text{ sin 2x dx }\]
\[\int x \cos^2 x\ dx\]
\[\int x^3 \cos x^2 dx\]
\[\int {cosec}^3 x\ dx\]
\[\int \cos^3 \sqrt{x}\ dx\]
\[\int e^x \sec x \left( 1 + \tan x \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
\[\int\sqrt{3 - x^2} \text{ dx}\]
\[\int\frac{x^2 + 1}{\left( 2x + 1 \right) \left( x^2 - 1 \right)} dx\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]
\[\int \cot^4 x\ dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int\sqrt{a^2 - x^2}\text{ dx }\]
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
