Advertisements
Advertisements
Question
\[\int x^2 \text{ cos x dx }\]
Sum
Advertisements
Solution
\[\int x^2 \text{ cos x dx }\]
` "Taking x"^2" as the first function and cos x as the second function " . `
\[ = x^2 \int\cos x dx - \int\left( \frac{d}{dx} x^2 \int\text{ cos x dx } \right)dx\]
\[ = x^2 \sin x - \int2x \text{ sin x dx }\]
\[ = x^2 \sin x - 2\left[ x\int\sin x - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin x dx } \right\}dx \right]\]
\[ = x^2 \sin x - 2\left[ - x\cos x + \int\text{ cos x dx } \right]\]
\[ = x^2 \sin x + 2x \cos x - 2 \sin x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int \cos^7 x \text{ dx } \]
Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{e^x}{\left( 1 + e^x \right)\left( 2 + e^x \right)} dx\]
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int\left( x + 1 \right) \text{ log x dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int x \sin^3 x\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int e^x \left( \tan x - \log \cos x \right) dx\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
\[\int x\sqrt{x^4 + 1} \text{ dx}\]
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int \tan^3 x\ dx\]
\[\int \tan^5 x\ dx\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
