Advertisements
Advertisements
प्रश्न
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
योग
Advertisements
उत्तर
∫ x . cos3 x2 sin x2 dx
Let x2 = t
⇒ 2x dx = dt
\[\Rightarrow \text{ x dx } = \frac{dt}{2}\]
\[Now, \int x . \cos^3 x^2 \sin x^2 dx\]
\[ = \frac{1}{2}\int \cos^3 t . \sin t . dt\]
\[\text{ Again let }\cos t = p\]
\[ \Rightarrow - \text{ sin t dt } = dp\]
\[ \Rightarrow \text{ sin t dt } = - dp\]
\[So, \frac{1}{2}\int \cos^3 t . \sin t . dt \]
\[ = - \frac{1}{2} p^3 \text{ dp }\]
\[ = - \frac{1}{2} \left( \frac{p^4}{4} \right) + C\]
\[ = - \frac{p^4}{8} + C\]
\[ = - \frac{\cos^4 t}{8} + C\]
\[ = - \frac{\cos^4 x^2}{8} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( x^e + e^x + e^e \right) dx\]
\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[\int\frac{1}{ x \text{log x } \text{log }\left( \text{log x }\right)} dx\]
\[\int\sqrt{1 + e^x} . e^x dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
\[\int \sec^4 2x \text{ dx }\]
\[\int \cot^5 x \text{ dx }\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int \log_{10} x\ dx\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \sqrt{x + 2}}\text{ dx}\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\sqrt{a^2 - x^2}\text{ dx }\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
