# Misc 15 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 15 Integrate the function cos^3x e^logsinx Let t = log (sin x) dt/dx=1/sinx cosx dt = 1/sinx cos〖x dx〗 "dt = " 1/e^t cos〖x dx〗 e^t dt=cos〖x dx〗 Substituting, ∫1▒〖e^logsinx cos^3〗x dx = ∫1▒〖e^t cos^2x 〗 cosx dx = ∫1▒e^t 〖cos〗^2 x e^t dt = ∫1▒e^2t 〖cos〗^2 x dt = ∫1▒〖e^2t (1-〖sin〗^2 x) dt 〗 = ∫1▒〖e^2t (1-e^2t ) dt〗 = ∫1▒(e^2t-e^4t ) dt = e^2t/2-e^4t/4+ C Putting value of e^t=sinx = ((〖sin〖x)〗〗^2)/2-(〖sin〗^4 x)/4+ C = (1 -〖 cos〗^2 x)/2- (1 -〖 cos〗^2 x)^2/4+ C = (1 -〖 cos〗^2 x)/4-((1 +〖 cos〗^4 -2〖cos〗^2 x)/4)+ C = 1/2-(〖cos〗^2 x)/2-1/4-(〖cos〗^4 x)/4+(〖cos〗^2 x)/2+ C = 1/4-(〖cos〗^4 x)/4+ C = (-〖cos〗^4 x)/4+ C_1 ∫1▒〖e^logsinx cos^3〗x = ∫1▒〖sin x cos^3〗〖x dx〗 = ∫1▒sin〖x 〖cos〗^2 x (cos〖x dx〗 〗 ) Let t = sin x dt/dx=cosx dt =cosx dx Substituting, = ∫1▒〖t 〖cos〗^2 〗 x dt = ∫1▒〖t (1-t^2 〗) dt = ∫1▒(t-t^3 ) dt = t^2/2-t^4/4+ C Putting value of t = ((〖sin〖x)〗〗^2)/2-(〖sin〗^4 x)/4+ C =(1 - 〖cos〗^2 x)/2- (1 - 〖cos〗^2 x)^2/4+ C = (1 -〖 cos〗^2 x)/4-((1 + 〖cos〗^4 -2〖cos〗^2 x)/4)+ C = 1/2-(〖cos〗^2 x)/2-1/4-(〖cos〗^4 x)/4+(〖cos〗^2 x)/2+ C = 1/4-(〖cos〗^4 x)/4+ C = (-〖cos〗^4 x)/4+ C_1

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Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.