Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\frac{\tan x}{\sec x + \tan x}dx\]
\[ = \int\frac{\tan x}{\left( \sec x + \tan x \right)} \times \left( \frac{\sec x - \tan x}{\sec x - \tan x} \right)dx\]
\[ = \int\frac{\tan x \left( \sec x - \tan x \right)}{\left( \sec^2 x - \tan^2 x \right)}dx\]
\[ = \int\left( \frac{\sec x \tan x - \tan^2 x}{1} \right)dx\]
\[ = \int\text{sec x }\text{tan x dx} - \int\left( se c^2 x - 1 \right)dx\]
\[ = \sec x - \tan x + x + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\int \tan^2 \left( 2x - 3 \right) dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
Evaluate the following integrals:
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
Write a value of
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
