Advertisements
Advertisements
प्रश्न
\[\int x \cos^3 x\ dx\]
योग
उत्तर
Let I=\[\int x \cos^3 x\ dx\]
\[\text{ As we know }, \]
\[\cos 3x = 4 \cos^3 x - 3\cos x\]
\[ \Rightarrow \cos^3 x = \frac{1}{4}\left( \cos 3x + 3\cos x \right)\]
\[\cos 3x = 4 \cos^3 x - 3\cos x\]
\[ \Rightarrow \cos^3 x = \frac{1}{4}\left( \cos 3x + 3\cos x \right)\]
\[\therefore I = \frac{1}{4}\int x . \left( \cos 3x + 3 \cos x \right)dx\]
\[ = \frac{1}{4}\int x_I . \text{ cos}_{II} \left( \text{ 3x }\right) dx + \frac{3}{4} \int x_I . \cos x_{II} \text{ dx }\]
\[ = \frac{1}{4}\left[ x . \int\text{ cos 3x dx }- \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos 3x dx }\right\}dx \right] + \frac{3}{4}\left[ x\int\cos x - \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos x dx }\right\}dx \right]\]
\[ = \frac{1}{4}\left[ x . \frac{\sin 3x}{3} - \int1 . \frac{\sin 3x}{3}dx \right] + \frac{3}{4}\left[ x\left( \sin x \right) - \int1 . \text{ sin x dx } \right]\]
\[ = \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3}{4}x \sin x + \frac{3}{4}\cos x + C\]
\[ = \frac{1}{4}\int x_I . \text{ cos}_{II} \left( \text{ 3x }\right) dx + \frac{3}{4} \int x_I . \cos x_{II} \text{ dx }\]
\[ = \frac{1}{4}\left[ x . \int\text{ cos 3x dx }- \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos 3x dx }\right\}dx \right] + \frac{3}{4}\left[ x\int\cos x - \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos x dx }\right\}dx \right]\]
\[ = \frac{1}{4}\left[ x . \frac{\sin 3x}{3} - \int1 . \frac{\sin 3x}{3}dx \right] + \frac{3}{4}\left[ x\left( \sin x \right) - \int1 . \text{ sin x dx } \right]\]
\[ = \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3}{4}x \sin x + \frac{3}{4}\cos x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int \sin^7 x \text{ dx }\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int \left( \log x \right)^2 \cdot x\ dx\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int x\sqrt{2x + 3} \text{ dx }\]
\[\int \cos^5 x\ dx\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
