substitution method 1/√1-tan²x. dx
Answers
Answer:
You have already been told about the useful identity
1+tan2x=1cos2x.
You may have seen this identity as
1+tan2x=sec2x.
There are slightly tricky things about taking square roots, but they are not a problem in the interval where you are working. We end up wanting to find ∫secxdx, or equivalently ∫dx/(cosx).
The strategy we will use is one that is useful when we are integrating a combination of powers of sinx and cosx, with one of the powers odd.
Rewrite the integral as
∫cosxcos2xdx=∫cosx1−sin2xdx.
Make the substitution u=sinx. Then du=cosxdx. So now our indefinite integral is
∫du1−u2.
It is more convenient to make the substitution in the "limits" of integration. When x=π/4, we have u=1/2–√ and when x=0, we have u=0, so we want
∫π/4u=0du1−u2.
Our integral is now in perfect shape for the method of "partial fractions." By going through the machinery of partial fractions, or by inspection, we have
11−u2=11−u2=1/21−u+1/21+u.
We have arrived at
∫1/2√0(1/21−u+1/21+u)du.
Finally, everything is easy. By substitution, or by inspection, we have
∫1/21−udu=−(1/2)ln(|1−u|)+Cand∫1/21+udu=(1/2)ln(|1+u|)+C.
(We don't have to worry about the constant of integration, since now we will be substituting the limits.) If we are in the mood, we can combine things, using the properties of logarithms, to conclude that the indefinite integral is
12ln(∣∣∣1+u1−u∣∣∣)+C.
A lot of work! But at least we got to practice a lot of techniques of integration. The integral ∫secxdx is one of the nastiest ones that one is likely to meet. Actually, it does show up in real applications.
A magic way: We want
∫secxdx.
Multiply "top" and "bottom" by secx+tanx. We get
∫sec2x+secxtanxsecx+tanxdx.
Let u=secx+tanx. Then from standard (?) differentiation formulas, we have
du=secxtanx+sec2x.
Note that this is more or less exactly what we have on top. So our indefinite integral simplifies to
∫duu=ln|u|+C
and it's over.
Answer:
Rewrite the integral as
Scosxcos2xdx=[cosx1-sin2xdx.
Make the substitution u-sinx. Then du-cosxdx. So now our indefinite integral is
Jdu1-u2.
It is more convenient to make the substitution in the "limits" of integration. When x=π/4, we have u-1/2-√ and when x=0, we have u=0, so we want
Sπ/4u=0du1-u2.
Our integral is now in perfect shape for the method of "partial fractions." By going through the machinery of partial fractions, or by inspection, we have
11-u2-11-u2=1/21-u+1/21+u.
We have arrived at
(1/2 √0(1/21-u+1/21+u)du.
Finally, everything is easy. By substitution, or by inspection, we have
[1/21-udu=-(1/2)In(|1-ul)