The integral of tan4x is:a)b)c)d)Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Begin by rewriting ∫tan⁴xdx as ∫tan²xtan²xdx.

Now we can apply the Pythagorean Identity, tan²x+1=sec²x, or tan²x=sec²x−1

∫tan²x tan²x dx = ∫(sec²x−1)tan²xdx

Distributing the tan²x:
= ∫sec²xtan²x − tan²xdx
Applying the sum rule:
= ∫sec²xtan²xdx − ∫tan²xdx

We'll evaluate these integrals one by one.

First Integral
This one is solved using a

Let u = tanx

Applying the substitution,

Because u = tanx,

Second Integral
Since we don't really know what ∫tan²xdx is by just looking at it, try applying the tan²x = sec²x−1

identity again:

∫tan²xdx = ∫(sec²x−1)dx

Using the sum rule, the integral boils down to:
∫sec²xdx − ∫1dx

The first of these, ∫sec²xdx, is just tanx + C.

The second one, the so-called "perfect integral", is simply x+C.

Putting it all together, we can say:
∫tan²xdx = tanx + C − x + C

And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan²xdx = tanx − x + C

Combining the two results, we have:
∫tan⁴xdx=∫sec²xtan²xdx−∫tan²xdx

=(tan³x/3 + C) − (tanx − x + C)

=tan³x/3 − tanx + x + C

Again, because C+C is a constant, we can join them into one C.

Answer

Begin by rewriting ∫tan⁴xdx as ∫tan²xtan²xdx.

Now we can apply the Pythagorean Identity, tan²x+1=sec²x, or tan²x=sec²x−1

∫tan²x tan²x dx = ∫(sec²x−1)tan²xdx

Distributing the tan²x:
= ∫sec²xtan²x − tan²xdx
Applying the sum rule:
= ∫sec²xtan²xdx − ∫tan²xdx

We'll evaluate these integrals one by one.

First Integral
This one is solved using a

Let u = tanx

Applying the substitution,

Because u = tanx,

Second Integral
Since we don't really know what ∫tan²xdx is by just looking at it, try applying the tan²x = sec²x−1

identity again:

∫tan²xdx = ∫(sec²x−1)dx

Using the sum rule, the integral boils down to:
∫sec²xdx − ∫1dx

The first of these, ∫sec²xdx, is just tanx + C.

The second one, the so-called "perfect integral", is simply x+C.

Putting it all together, we can say:
∫tan²xdx = tanx + C − x + C

And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan²xdx = tanx − x + C

Combining the two results, we have:
∫tan⁴xdx=∫sec²xtan²xdx−∫tan²xdx

=(tan³x/3 + C) − (tanx − x + C)

=tan³x/3 − tanx + x + C

Again, because C+C is a constant, we can join them into one C.

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