Begin by using the sum rule to break #intcos^2x-tan^3x+sinxdx# into:
#intcos^2xdx-inttan^3xdx+intsinxdx#
First Integral
Since we don't know what #intcos^2xdx# is immediately, we have to tap into our trig knowledge. I hope you remember (or you can find, in your calculus textbook perhaps), the formula for #cos^2x#:
#cos^2x=1/2(1+cos2x)#
Try to integrate this:
#intcos^2xdx=int1/2(1+cos2x)dx#
#color(white)(XX)=1/2(int1dx+intcos2xdx)#
#color(white)(XX)=1/2(x+C_1+(sin2x)/2+C_2)#
#color(white)(XX)=(2x+sin2x)/4+C#
Second Integral
Again, we don't know what #inttan^3xdx# is, so we use our knowledge of trig (i.e. Pythagorean Identities) to solve it:
#inttan^3xdx=inttanxtan^2xdx#
#color(white)(XX)=inttanx(sec^2x-1)dx#
#color(white)(XX)=inttanxsec^2xdx-inttanxdx#
For the first of these, let #u=tanx# and #(du)/dx=sec^2x#; that makes #du=sec^2xdx#:
#inttanxsec^2xdx=intudu=u^2/2+C_1#
Because #u=tanx#, #inttanxsec^2xdx=(tanx)/2+C_1#.
The second integral here, #inttanxdx#, is simply #lnabs(cosx)+C_2#. Combining these two results, we see #inttan^3x=tan^2x/2+lnabs(cosx)+C#.
Third Integral
Thankfully, #intsinxdx# is the most simple of all: it's just #-cosx+C#.
Now put all of it together:
- #intcos^2xdx=(2x+sin2x)/4+C_1#
- #inttan^3xdx=tan^2x/2+lnabs(cosx)+C_2#
- #intsinxdx=-cosx+C_3#
#intcos^2xdx-inttan^3xdx+intsinxdx=(2x+sin2x)/4+C_1-tan^2x/2-lnabs(cosx)+C_2-cosx+C_3=(2x+sin2x)/4-tan^2x/2-lnabs(cosx)-cosx+C#
We are being asked for the value of #C# if #f(pi/6)=1#:
#1=(2(pi/6)+sin2(pi/6))/4-tan^2(pi/6)/2-lnabs(cos(pi/6))-cos(pi/6)+C#
#1=(pi/3+sqrt(3)/2)/4-(1/3)/2-lnabs(sqrt(3)/2)-sqrt(3)/2+C#
#1=(2pi+3sqrt(3))/24-1/6-lnabs(sqrt(3)/2)-sqrt(3)/2+C#
#1=(2pi+3sqrt(3)-4)/24-lnabs(sqrt(3)/2)-sqrt(3)/2+C#
#C=-(2pi+3sqrt(3)-4)/24+lnabs(sqrt(3)/2)+sqrt(3)/2+1#
#C=(28-2pi-3sqrt(3))/24+lnabs(sqrt(3)/2)+sqrt(3)/2#
Thus #f(x)=(2x+sin2x)/4-tan^2x/2-lnabs(cosx)-cosx+(28-2pi-3sqrt(3))/24+lnabs(sqrt(3)/2)+sqrt(3)/2#
