Advertisements
Advertisements
Question
\[\int x^2 \text{ cos x dx }\]
Sum
Solution
\[\int x^2 \text{ cos x dx }\]
` " Taking x"^2" as the first function and cos x as the second function . " `
\[ = x^2 \int\text{ cos x dx } - \int\left\{ \frac{d}{dx}\left( x^2 \right)\int\text{ cos x dx }\right\}dx\]
\[ = x^2 \sin x - \int2x \text{ sin x dx }\]
\[ = x^2 \sin x - 2\left[ x\int\sin x - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin x dx }\right\}dx \right]\]
\[ = x^2 \sin x + 2x\cos x - 2\int\text{ cos x dx }\]
\[ = x^2 \sin x + 2x \cos x - 2 \sin x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int\frac{1}{\sin x \cos^3 x} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{1 + x + x^2 + x^3} dx\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
If \[\int\frac{2^{1/x}}{x^2} dx = k 2^{1/x} + C,\] then k is equal to
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
