This problem was assigned to a student I'm tutoring as extra credit:
∫x(1+sinx)2dx
I have two potential methods in mind, though I suspect the first, which employs the Weierstrass tangent half-angle substitution, goes beyond the student's curriculum, while the second probably falls within what he's learned (or at least in my opinion, what he should absolutely know). Both methods are rather lengthy, and, while I'm convinced the "extra credit" part about this problem is correlated with that, I can't help but wonder if there are more straightforward methods to do it.
I outline my suggestions below.
Method 1:
Integrate by parts to get
∫x(1+sinx)2dx=uv−∫vduwhere u=x and v=∫dx(1+sinx)2.Compute v using the Weierstrass substitution t=tanx2, which yields
−2+3tanx2+3tan2x23(1+tanx2)3x+∫2+3tanx2+3tan2x23(1+tanx2)3dxCompute the remaining integral by expanding like so:
∫dx(1+tanx2)2+∫1+tan2x2(1+tanx2)2dx−43∫dx(1+tanx2)3and with appropriate substitutions (Weierstrass and w=1+tanx2) rewrite as
∫2(1+t)2(1+t2)dt+∫2w2dw−43∫2(1+t)3(1+t2)dtDecompose into partial fractions, etc.
Method 2:
Let x=π2−y and rewrite the integral as
∫x(1+sinx)2dx=∫y−π2(1+cosy)2dyRecalling the half-angle identity for cosine, we can rewrite the integrand as
∫(y4−π8)sec4y2dy=∫(y−π4)sec4ydyIntegrate by parts
∫(y−π4)sec4ydy=uv−∫vduthis time with u=y−π4 and v=∫sec4ydy.Compute v with a reduction formula or the expansion sec4y=sec2y(1+tan2y), etc.
Answer
If you let u=x,v=∫1(1+sinx)2dx=∫1(sinx2+cosx2)2dx=∫1(√2cos(x2−π4))2dx
=12∫sec2(x2−π4)dx=tan(x2−π4)+C, this gives
∫1(1+sinx)2dx=uv−∫vdu=xtan(x2−π4)−[−2ln|cos(x2−π4)|]+C
=xtan(x2−π4)+2ln|cos(x2−π4)|+C
No comments:
Post a Comment