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