Advertisements
Advertisements
Question
\[\int \cos^2 \frac{x}{2} dx\]
Sum
Advertisements
Solution
\[\int \cos^2 \frac{x}{2} dx\]
\[ = \int\left( \frac{1 + \cos x}{2} \right)dx \left[ \therefore \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 + \cos x \right)dx\]
\[ = \frac{1}{2}\left[ x + \sin x \right] + C\]
\[ = \int\left( \frac{1 + \cos x}{2} \right)dx \left[ \therefore \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 + \cos x \right)dx\]
\[ = \frac{1}{2}\left[ x + \sin x \right] + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]
\[\int x^3 \cos x^4 dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int \sec^4 2x \text{ dx }\]
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int\frac{1}{x^2 + 6x + 13} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
`int 1/(cos x - sin x)dx`
\[\int x e^x \text{ dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int e^x \left( \cos x - \sin x \right) dx\]
\[\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{1}{1 + x + x^2 + x^3} \text{ dx }\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
