हिंदी

∫ Cos X √ 4 + Sin 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{\cos x}{\sqrt{4 + \sin^2 x}} dx\]
योग

उत्तर

 

` ∫   { cos x  dx}/{\sqrt{4 + sin^2 x}} `
\[\text{ let }\sin x = t\]
\[ \Rightarrow \text{ cos x dx }= dt\]
Now, ` ∫   { cos x  dx}/{\sqrt{4 + sin^2 x}} `
\[ = \int\frac{dt}{\sqrt{2^2 + t^2}}\]
\[ = \text{ log } \left| t + \sqrt{4 + t^2} \right| + C\]
\[ = \text{ log } \left| \sin x + \sqrt{4 + \sin^2 x} \right| + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.18 [पृष्ठ ९९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.18 | Q 4 | पृष्ठ ९९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]

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

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

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


\[\int\sin x\sqrt{1 + \cos 2x} dx\]

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

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

\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]

\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


\[\int\frac{a}{b + c e^x} dx\]

\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]

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

\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]

` ∫  tan 2x tan 3x  tan 5x    dx  `

\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]

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

` ∫  tan^5 x   sec ^4 x   dx `

\[\int \cos^5 x \text{ dx }\]

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

\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{  dx }\]

`int"x"^"n"."log"  "x"  "dx"`

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

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

\[\int x \cos^3 x\ dx\]

\[\int e^x \sec x \left( 1 + \tan x \right) dx\]

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

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

\[\int\frac{1}{7 + 5 \cos x} dx =\]

\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

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

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

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

\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]


\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×