Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
योग
Advertisements
उत्तर
\[\text{ Let I }= \int \frac{1}{\cos x\left( \sin x + 2 \cos x \right)}dx\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x}{\frac{\cos x}{\cos x} \times \left( \frac{\sin x + 2 \cos x}{\cos x} \right)}dx\]
\[ = \int \frac{\sec^2 x}{\left( \tan x + 2 \right)}dx\]
\[\text{ Let tan x } + 2 = t\]
\[ \Rightarrow \sec^2 x \text{ dx } = dt\]
\[ \therefore I = \int \frac{dt}{t}\]
\[ = \text{ ln } \left| t \right| + C\]
\[ = \text{ ln } \left| \tan x + 2 \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
` ∫ cos 3x cos 4x` dx
\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{x^4 + 1}{x^2 + 1} dx\]
\[\int\frac{1}{1 + x - x^2} \text{ dx }\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int x e^x \text{ dx }\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
