Advertisements
Advertisements
Question
\[\int \sin^2 \frac{x}{2} dx\]
Sum
Advertisements
Solution
\[\int \sin^2 \frac{x}{2} dx\]
\[ = \int\left( \frac{1 - \cos x}{2} \right)dx \left[ \therefore \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos x \right)dx\]
\[ = \frac{1}{2}\left[ x - \sin x \right] + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\int\frac{x}{\left( x^2 - a^2 \right) \left( x^2 - b^2 \right)} dx\]
\[\int\frac{x}{\left( x - 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]
\[\int \sec^6 x\ dx\]
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
\[\int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int x^2 \tan^{- 1} x\ dx\]
