Evaluate the following integral ∫〖〖tan〗^2 x〗 〖sec〗^4 x dx
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total answers (1)
Mathematics
∫〖〖tan〗^2 x〗 〖sec〗^4 x dx=∫〖〖tan〗^2 x〗 〖sec〗^2 x 〖sec〗^2 x dx
need an explanation for this answer? contact us directly to get an explanation for this answer=∫〖〖tan〗^2 x(〖tan〗^2 x+1) 〖sec〗^2 x 〗 dx
If we let u=tanx ,then du= 〖sec〗^2 x dx,and:
∫〖〖tan〗^2 x〗 〖sec〗^4 x dx=∫〖u^2 (u^2+1) 〗 du
=∫(u^4+u^2 ) du
=1/5 u^5+1/3 u^3+c
=1/5 〖tan〗^5 x+1/3 〖tan〗^3 x+c