int sin(x)/cos^3(x) dx∫sin(x)cos3(x)dx
Let us use the substitution method.
We can see that the derivative of cos(x)cos(x) is -sin(x)−sin(x)
Let cos(x)= ucos(x)=u
Differentiating with respect to xx we get
-sin(x)dx = du−sin(x)dx=du
sin(x)dx = -dusin(x)dx=−du
Our integral becomes
int (-du)/u^3∫−duu3
=-int u^-3 du=−∫u−3du
=-(u^(-3+1)/(-3+1))+C=−(u−3+1−3+1)+C
=-u^-2/-2+C=−u−2−2+C
=1/(2u^2)+C=12u2+C
Substituting back
=1/(2cos^2(x))+C=12cos2(x)+C
We can also write this as
=1/2 sec^2(x)+C=12sec2(x)+C
Note this can also be written as 1/2 tan^2(x) +C12tan2(x)+C
If you are wondering why? It is nothing but replacing sec^2(x)sec2(x) as 1+tan^2(x)1+tan2(x). Let me show how.
=1/2(1+tan^2)x+C=12(1+tan2)x+C
=1/2+1/2(tan^2(x)+C=12+12(tan2(x)+C
=1/2tan^2(x)+C=12tan2(x)+C since CC is a constant and 1/2+C12+C would be a constant too. We can write it as CC itself.
For integration involving trigonometric functions there are usually more than one way of giving an answer so just watch out.
