Advertisements
Advertisements
प्रश्न
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
योग
Advertisements
उत्तर
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[ = \int\left( \frac{1 - \cos \left( 4x + 10 \right)}{2} \right)dx \left[ \therefore \sin^2 A = \frac{1 - \cos2A}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos \left( 4x + 10 \right) \right)dx\]
\[ = \frac{1}{2}\left[ x - \frac{\sin \left( 4x + 10 \right)}{4} \right] + C\]
\[ = \frac{1}{2}x - \frac{\sin \left( 4x + 10 \right)}{8} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^{1/3}} \right)dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
Integrate the following integrals:
\[\int\text{sin 2x sin 4x sin 6x dx} \]
Integrate the following integrals:
\[\int\text { sin x cos 2x sin 3x dx}\]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} dx\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{x^2 - 10x + 34} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{x + 1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\frac{1}{1 - \cos x - \sin x} dx =\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int \left( \sin^{- 1} x \right)^3 dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
