Advertisements
Advertisements
प्रश्न
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
विकल्प
log (3 + 4 cos2 x) + C
- \[\frac{1}{2 \sqrt{3}} \tan^{- 1} \left( \frac{\cos x}{\sqrt{3}} \right) + C\]
- \[- \frac{1}{2 \sqrt{3}} \tan^{- 1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C\]
- \[\frac{1}{2 \sqrt{3}} \tan^{- 1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C\]
MCQ
उत्तर
\[- \frac{1}{2 \sqrt{3}} \tan^{- 1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C\]
\[\text{Let }I = \int\frac{\sin x}{3 + 4 \cos^2 x}dx\]
\[\text{Putting }\cos x = t\]
\[ \Rightarrow - \sin x dx = dt\]
\[ \therefore I = \int\frac{- dt}{3 + 4 t^2}\]
\[ = \frac{1}{4}\int\frac{- dt}{t^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{- 1}{4} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \left( \frac{t \times 2}{\sqrt{3}} \right) + C .............\left( \because \int\frac{1}{x^2 + a^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = - \frac{1}{2\sqrt{3}} \tan^{- 1} \left( \frac{2 t}{\sqrt{3}} \right) + C\]
\[ = - \frac{1}{2\sqrt{3}} \tan^{- 1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C .............\left( \because t = \cos x \right)\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int \sin^2 \frac{x}{2} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
` ∫ tan^5 x dx `
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]
` ∫ x tan ^2 x dx
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int e^x \sec x \left( 1 + \tan x \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int x\sqrt{x^4 + 1} \text{ dx}\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]
\[\int\frac{1}{1 + x + x^2 + x^3} dx\]
\[\int\frac{1}{\sin x \left( 3 + 2 \cos x \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int \sec^4 x\ dx\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
