Integration using standard results ?

Question

I was given some integration questions to do and there were a couple which i didnt know how to go about solving =s

Integrate:

1) [(secx)^2 (1+ (e^x)(cosx)^2) ] dx
2) [(1 + cosx)/(sinx)^2 + (1+x)/ (x^2)] dx

thank you so much if you can help me with these showing the working. hopefully when i get help with the above ill be able to work out the rest

thanks!

Answer 1

1) Integral [ (sec^2(x)) (1 + e^x cos^2(x) ) ] dx

One thing you can do is distribute sec^2(x).

Integral ( [ sec^2(x) + e^x cos^2(x) sec^2(x) ] dx )

It's worthy to note that cos^2(x) and sec^2(x) are reciprocals of each other; for that reason, they cancel to 1. This gives us

Integral ( [sec^2(x) + e^x] dx )

Now, these two are known integrals, and we can integrate normally.

tan(x) + e^x + C

2) Integral [ (1 + cos(x))/sin^2(x) + (1 + x)/x^2 ) ] dx

Let's split each fraction up into two.

Integral [ 1/sin^2(x) + cos(x)/sin^2(x) + 1/x^2 + x/x^2 ] dx

which of courses reduces as

Integral [csc^2(x) + (cos(x)/sin(x))(1/sin(x)) + x^(-2) + 1/x] dx

{Side note: I also decomposed cos(x)/sin^2(x) as a product of two fractions; cos(x)/sin(x) and 1/sin(x)}

Integral [csc^2(x) + cot(x)csc(x) + x^(-2) + 1/x] dx

Keep in mind that csc^2(x) is *almost* a known derivative, in the sense that d/dx cot(x) = -csc^2(x). For that reason, since it is missing a negative sign (which is a constant), all we have to do is offset the integral with a minus sign as well.

-cot(x) + Integral (cot(x)csc(x) + x^(-2) + 1/x) dx

The same thing goes with cot(x)csc(x); it's ALMOST a known derivative, because d/dx csc(x) = -csc(x)cot(x). Offset this with a negative as well.

-cot(x) - csc(x) + Integral (x^(-2) + 1/x) dx

For x^(-2), just use the reverse power rule, so we get (1/(-1))x^(-1), or put simply, -1/x.

-cot(x) - csc(x) - 1/x + Integral (1/x dx )

Now, this final integral is simple enough; it's ln|x|.

-cot(x) - csc(x) - 1/x + ln|x| + C