हिंदी

∫ ( Tan X + Cot X ) 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \left( \tan x + \cot x \right)^2 dx\]
योग
Advertisements

उत्तर

\[\int \left( \tan x + \cot x \right)^2 \]
\[ = \int\left( \tan^2 x + \cot^2 x + 2 \tan x \cot x \right)dx\]
\[ = \int\left( \tan^2 x + \cot^2 x + 2 \right)dx\]
\[ = \int\left[ \left( \sec^2 x - 1 \right) + \left( {cosec}^2 x - 1 \right) + 2 \right]dx\]
\[ = \int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[ = \tan x - \cot x + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.02 | Q 25 | पृष्ठ १५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

`int{sqrtx(ax^2+bx+c)}dx`

\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]

\[\int\frac{1 - \sin x}{x + \cos x} dx\]

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]

\[\int\frac{e^{2x}}{1 + e^x} dx\]

\[\int\frac{1}{\left( x + 1 \right)\left( x^2 + 2x + 2 \right)} dx\]

` ∫  tan^5 x   sec ^4 x   dx `

` ∫  sec^6   x  tan    x   dx `

\[\int {cosec}^4  \text{ 3x } \text{ dx } \]

\[\int\frac{dx}{e^x + e^{- x}}\]

\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]

\[\int\frac{x + 1}{x^2 + x + 3} dx\]

\[\int\frac{2x}{2 + x - x^2} \text{ dx }\]

\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]

\[\int\frac{1}{5 + 4 \cos x} dx\]

\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

\[\int x \text{ sin 2x dx }\]

\[\int x \sin x \cos x\ dx\]

 


\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]

\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]

\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]

\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]

\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]

\[\int\frac{x + 1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]

\[\int \cot^5 x\ dx\]

\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]

\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]

\[\int \sin^{- 1} \sqrt{x}\ dx\]

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]

\[\int \sin^3  \left( 2x + 1 \right)  \text{dx}\]


\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×