MHB Integrating Higher Powers of Tangent: A Shortcut Using Trigonometric Identities

[karush]

7.2.6a use $\tan^2 x= \sec^2 x-1$ to evaluate
$$\displaystyle I_{6a}=\int\tan^4 x \, dx$$
well my first inclinations is to.
$$\int(\sec^2 x-1)^2\, dx$$
then expand
$$\int (\sec^4 x - 2 \sec^2 x +1)$$
ok not sure if this is the right direction

W|A returned this:
View attachment 9281

 

[MarkFL]

I would write:

$$I=\int \tan^2(x)(\sec^2(x)-1)\,dx=\int \tan^2(x)\sec^2(x)\,dx-\int \tan^2(x)\,dx$$

The first integral is straightforward ad for the second, apply the identity again:

$$I=\frac{1}{3}\tan^3(x)-\int \sec^2(x)-1\,dx$$

$$I=\frac{1}{3}\tan^3(x)-\tan(x)+x+C$$

Now, let's factor:

$$I=\frac{1}{3}\tan(x)\left(\tan^2(x)-3\right)+x+C$$

Apply the identity again:

$$I=\frac{1}{3}\tan(x)\left(\sec^2(x)-4\right)+x+C$$

 

[karush]

that is cool..

so many examples went for the U substitution optionalways seem to get the best help here

Mahalo

 

[topsquark]

I would write:

$$I=\int \tan^2(x)(\sec^2(x)-1)\,dx=\int \tan^2(x)\sec^2(x)\,dx-\int \tan^2(x)\,dx$$

The first integral is straightforward ad for the second, apply the identity again:

$$I=\frac{1}{3}\tan^3(x)-\int \sec^2(x)-1\,dx$$

$$I=\frac{1}{3}\tan^3(x)-\tan(x)+x+C$$

Now, let's factor:

$$I=\frac{1}{3}\tan(x)\left(\tan^2(x)-3\right)+x+C$$

Apply the identity again:

$$I=\frac{1}{3}\tan(x)\left(\sec^2(x)-4\right)+x+C$$

Nicely done! (Bow)

-Dan