Can You Verify the Integral of Secant Squared Over (1+Tangent)^3?

[karush]

$$\int\frac{\sec^2 \left({t}\right)}{\left(1+\tan\left({t}\right)\right)^3}dt$$

$$u=1+\tan\left({t}\right)\ du=\sec^2\left({t}\right) dt$$

So far ok?

 

[I like Serena]

$$\int\frac{\sec^2 \left({t}\right)}{\left(1+\tan\left({t}\right)\right)^3}dt$$

$$u=1+\tan\left({t}\right)\ du=\sec^2\left({t}\right) dt$$

So far ok?

Yep. (Nod)

 

[karush]

$$\int\frac{1}{{u}^{3}}du =-\frac{1}{{2u}^{2}}=\frac{1}{2 (1+\tan\left({t}\right)) ^2 }+C$$

I hope the TI gave a different answer?

 

[I like Serena]

$$\int\frac{1}{{u}^{3}}du =-\frac{1}{{2u}^{2}}=\frac{1}{2 (1+\tan\left({t}\right)) ^2 }+C$$

I hope the TI gave a different answer?

You've lost a minus sign in the final answer, but otherwise we can verify that if we take the derivative we get indeed the original integral.