Advertisements
Advertisements
प्रश्न
\[\int x^3 \cos x^2 dx\]
बेरीज
उत्तर
\[\int x^3 \cos x^2 \text{ dx }\]
\[\text{ Let x}^2 = t \]
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{2x}\]
\[ = \frac{1}{2}\left[ \int t \text{ cos t dt } \right]\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = \frac{1}{2}\left[ t\sin t - \int\text{ sin t dt } \right]\]
\[ = \frac{1}{2}\left[ t\sin t + \cos t \right] . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) } \]
\[ = \frac{x^2 \sin x^2}{2} + \frac{\cos x^2}{2} + c\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int \cos^2 \frac{x}{2} dx\]
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int \cos^7 x \text{ dx } \]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{2 x^2 - x - 1} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
`int 1/(cos x - sin x)dx`
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{x^4 - 1} dx\]
\[\int \tan^5 x\ dx\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
