Advertisements
Advertisements
प्रश्न
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
योग
Advertisements
उत्तर
\[\text{ Let I } = \int e^{2x} \left( - \sin x + 2\cos x \right)dx\]
\[\text{ Put e}^{2x} \cos x = t\]
\[\text{ Diff both sides w . r . t x }\]
\[\left[ 2 e^{2x} \cos x + e^{2x} \times \left( - \sin x \right) \right]dx = dt\]
` ∴ \text{ I } = \int dt `
\[ = t + C\]
\[ = e^{2x} \cos x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\ ∫ x \text{ e}^{x^2} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int \log_{10} x\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
` ∫ x tan ^2 x dx
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]
