Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 30

int_0^(pi/4) tan^4(t) dt
Express the integrand in factor form.
= int _0^(pi/4) tan^2(t) *tan^2(t) dt
Plug-in the trigonometric identity tan^2(t) = sec^2(t)-1 to one of the factors.
= int_0^(pi/4) tan^2(t)*(sec^2(t)-1) dt
= int_0^(pi/4) (tan^2(t)sec^2(t)-tan^2(t)) dt
= int _0^(pi/4) tan^2(t)sec^2(t) dt - int_0^(pi/4)tan^2(t) dt
Plug-in again the trigonometric identity tan^2(t) =sec^2(t) - 1 to the second integral.
= int_0^(pi/4) tan^2(t)sec^2(t)dt- int_0^(pi/4) (sec^2(t) - 1) dt
= int_0^(pi/4) tan^2(t) sec^2(t) dt - int_0^(pi/4) sec^2(t)dt + int_0^(pi/4) dt
For the first integral, apply the u-substitution method.
u=tan (t)
du = sec^2(t) dt
>> int tan^2(t) sec^2(t)dt = int u^2du=u^3/3 = (tan^3(t))/3
For the second integral, apply the formula int sec^2 x dx = tanx .
>> int sec^2(t)dt = tan (t)
And for the third integral, apply the formula int adx = ax .
gtgt int dt = t
So the integral becomes:
int_0^(pi/4) tan^2(t) sec^2(t) dt - int_0^(pi/4) sec^2(t)dt + int_0^(pi/4) dt
= ((tan^3(t))/3 -tan(t) + t) |_0^(pi/4)
=((tan^3(pi/4))/3 - tan(pi/4) + pi/4) - ((tan^3(0))/3 - tan(0) + 0)
= (1/3 - 1 + pi/4) - 0
=pi/4 - 2/3
Therefore, int_0^(pi/4) tan^4(t) dt = pi/4 - 2/3 .