Advertisements
Advertisements
प्रश्न
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
बेरीज
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
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int \tan^{- 1} \left( \frac{\sin 2x}{1 + \cos 2x} \right) dx\]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{1 - \sin 2x}{x + \cos^2 x} dx\]
\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\ ∫ x \text{ e}^{x^2} dx\]
` ∫ tan^3 x sec^2 x dx `
\[\int \sin^3 x \cos^6 x \text{ dx }\]
` ∫ {1}/{a^2 x^2- b^2}dx`
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int x^2 \text{ cos x dx }\]
\[\int e^x \left( \cos x - \sin x \right) dx\]
\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]
\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
