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If \int \sec ^{5} x\tan ^{3 }xdx= \frac{\sec ^{A}x}{A}+\frac{\sec ^{B}}{D}+C   Then the value of A+B+D=  

  • Option 1)

    7

  • Option 2)

    12

  • Option 3)

    17

  • Option 4)

    none of these

 

Answers (1)

As we have learned

Special type of indefinite integration -

Integrals of the form :

(tan^{m}x)(sec^{n}x)

- wherein

Use 

sec^{2}x-tan^{2}x=1

 

\int \sec^{5}x \tan ^{3}x dx

= \int \sec^{4}x (\sec ^{2}x-1) \sec x \tan x dx

(put t = sec x ; dt = sec x tan x dx) 

 

= \frac{\sec^{7}x}{7}+\frac{\sec ^{5}x}{5}+ C  or 

= \frac{\sec^{7}x}{7}+\frac{\sec ^{5}x}{-5}+ C

A= 7, B =5 , D= -5 

A+B+D= 7

 

 


Option 1)

7

This is correct

Option 2)

12

This is incorrect

Option 3)

17

This is incorrect

Option 4)

none of these

This is incorrect

Posted by

subam

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