Chapter 7 Class 12 Integrals

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Ex 7.6, 13 - Chapter 7 Class 12 Integrals (Important Question)

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Integration of tan inverse x - Ex 7.6, 13 - Chapter 7 Class 12

Ex 7.6, 13 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 13 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.6, 13 Integrate the function - tan^(āˆ’1) š‘„ āˆ«1ā–’怖" " tan^(āˆ’1) š‘„" " 怗 .š‘‘š‘„=āˆ«1ā–’怖(tan^(āˆ’1) š‘„) 1.š‘‘š‘„ " " 怗 = tan^(āˆ’1) š‘„āˆ«1ā–’怖1 .怗 š‘‘š‘„āˆ’āˆ«1ā–’(š‘‘(tan^(āˆ’1)ā”š‘„ )/š‘‘š‘„ āˆ«1ā–’怖1 .š‘‘š‘„怗) š‘‘š‘„ = tan^(āˆ’1) š‘„ (š‘„)āˆ’āˆ«1ā–’1/(1 + š‘„^2 ) . š‘„ . š‘‘š‘„ = š‘„ tan^(āˆ’1) š‘„āˆ’āˆ«1ā–’š‘„/(1 + š‘„^2 ) . š‘‘š‘„ Now we know that āˆ«1ā–’怖š‘“(š‘„) š‘”ā”(š‘„) 怗 š‘‘š‘„=š‘“(š‘„) āˆ«1ā–’š‘”(š‘„) š‘‘š‘„āˆ’āˆ«1ā–’(š‘“ā€²(š‘„)āˆ«1ā–’š‘”(š‘„) š‘‘š‘„) š‘‘š‘„ Putting f(x) = tanā€“1 x and g(x) = 1 Solving I1 I1 = āˆ«1ā–’š‘„/(1 + š‘„^2 ) . š‘‘š‘„" " Let 1 + š‘„^2=š‘” Differentiating both sides š‘¤.š‘Ÿ.š‘”.š‘„ 0 + 2š‘„=š‘‘š‘”/š‘‘š‘„ š‘‘š‘„=š‘‘š‘”/2š‘„ Our equation becomes I1 = āˆ«1ā–’š‘„/(1 + š‘„^2 ) . š‘‘š‘„" " Putting the value of (1+š‘„^2 ) = t and š‘‘š‘„ = š‘‘š‘”/( 2š‘„) , we get I1 = āˆ«1ā–’š‘„/š‘” . š‘‘š‘”/2š‘„ I1 = 1/2 āˆ«1ā–’1/š‘” . š‘‘š‘” I1 = 1/2 logā”怖 |š‘”|怗+š¶1 I1 = 1/2 logā”怖 |1+š‘„^2 |怗+š¶1 Putting the value of I1 in (1) , āˆ«1ā–’怖" " tan^(āˆ’1) š‘„" " 怗 .š‘‘š‘„=š‘„ tan^(āˆ’1) š‘„āˆ’āˆ«1ā–’š‘„/(1 + š‘„^2 ) . š‘‘š‘„ =š‘„ tan^(āˆ’1) š‘„āˆ’(1/2 怖log 怗ā”|1+š‘„^2 |+š¶1) =š‘„ tan^(āˆ’1) š‘„āˆ’1/2 怖log 怗ā”|1+š‘„^2 |āˆ’š¶1 =š’™ 怖š’•š’‚š’ć€—^(āˆ’šŸ) š’™āˆ’šŸ/šŸ 怖š’š’š’ˆ 怗ā”(šŸ+š’™^šŸ )+š‘Ŗ

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