Misc 15 - Integrate cos3 x elog sinx - Chapter 7 NCERT

Misc 15 Integrate the function cos^3⁡x e^log⁡sin⁡x Let t = log (sin x) dt/dx=1/sin⁡x cos⁡x dt = 1/sin⁡x cos⁡〖x dx〗 "dt = " 1/e^t cos⁡〖x dx〗 e^t dt=cos⁡〖x dx〗 Substituting, ∫1▒〖e^log⁡sin⁡x cos^3〗⁡x dx = ∫1▒〖e^t cos^2⁡x 〗 cos⁡x 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=sin⁡x = ((〖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^log⁡sin⁡x 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=cos⁡x dt =cos⁡x 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

Misc 15 - Chapter 7 Class 12 Integrals