Saturday, June 17, 2017

`int_0^(pi/4) tan^4(t) dt` Evaluate the integral

`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`


...

`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` .

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