हिंदी

∫ Sec X C O S E C X Log ( Tan X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

` Note: Here ,  we   are "  considering " log x  as   log_e x` .
      Let I = ` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx
`  "Putting"  "log" \ tan x = t `
\[ \Rightarrow \frac{\sec^2 x}{\tan x} = \frac{dt}{dx}\]
\[ \Rightarrow \text{sec x cosec x dx} = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log }\left| \text{t }\right| + C\]
\[ = \text{log} \left| \text{log} \left( \tan x \right) \right| + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.08 | Q 15 | पृष्ठ ४७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


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

\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]

\[\int \tan^3 \text{2x sec 2x dx}\]

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

` ∫      tan^5    x   dx `


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

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

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

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

\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]

\[\int\frac{x^3}{x^4 + x^2 + 1}dx\]

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

\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]

\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]

\[\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }\]

\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]

\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]

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

\[\int e^\sqrt{x} \text{ dx }\]

\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]

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

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

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


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

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\] 

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

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

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

\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]

\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×