Question

I=tan^-1(sec3x+tan3x) dx
pls ans this integration
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Answer 1

Answer:

We have

... sec 3x + tan 3x

= ( 1 + sin 3x ) / cos 3x

= ( cos 3x/2 + sin 3x/2 )² / [ ( cos 3x/2 + sin 3x/2)( cos 3x/2 - sin 3x/2 ) ]

= ( cos 3x/2 + sin 3x/2 ) / ( cos 3x/2 - sin 3x/2 )

= ( 1 + tan 3x/2 ) / ( 1 - tan 3x/2 )

= ( tan π/4 + tan 3x/2 ) / ( 1 - tan π/4 tan 3x/2 )

= tan (π/4 + 3x/2) ...................................................... (1)

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Then, the required integral is

= ∫ tanֿ¹ [ tan (π/4 + 3x/2) ] dx

= ∫ (π/4 + 3x/2) dx

= (π/4) ∫ dx + (3/2) ∫ x dx

= (π/4)x + (3/2)( x² / 2 ) + C

= (1/4) ( πx + 3x² ) + C ..................... Ans.

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Happy To Help ! ^-^

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Answer 2

Step-by-step explanation:

... sec 3x + tan 3x

= ( 1 + sin 3x ) / cos 3x

= ( cos 3x/2 + sin 3x/2 )² / [ ( cos 3x/2 + sin 3x/2)( cos 3x/2 - sin 3x/2 ) ]

= ( cos 3x/2 + sin 3x/2 ) / ( cos 3x/2 - sin 3x/2 )

= ( 1 + tan 3x/2 ) / ( 1 - tan 3x/2 )

= ( tan π/4 + tan 3x/2 ) / ( 1 - tan π/4 tan 3x/2 )

= tan (π/4 + 3x/2) ........................................... (1)

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Then, the required integral is

= ∫ tanֿ¹ [ tan (π/4 + 3x/2) ] dx

= ∫ (π/4 + 3x/2) dx

= (π/4) ∫ dx + (3/2) ∫ x dx

= (π/4)x + (3/2)( x² / 2 ) + C

= (1/4) ( πx + 3x² ) + C ..................... Ans.