Worksheet: Integration - Integrals Yielding Inverse Trigonometric ... - Free Printable

Let’s solve each integral one by one. I’ll go step by step, check my work, and then give the final answers.

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a) ∫ 1 / (x² + 25) dx

This is a standard form:
∫ 1 / (x² + a²) dx = (1/a) arctan(x/a) + C

Here, a² = 25 → a = 5

So:

→ (1/5) arctan(x/5) + C

✔ Checked — correct.

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b) ∫ x / (x² + 4) dx

Use substitution.

Let u = x² + 4 → du = 2x dx → (1/2) du = x dx

So integral becomes:

∫ (1/u) * (1/2) du = (1/2) ln|u| + C = (1/2) ln(x² + 4) + C

(Note: x² + 4 is always positive, so absolute value not needed.)

✔ Checked — correct.

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c) ∫ 9x / (4x² - 16) dx

First, factor denominator: 4x² - 16 = 4(x² - 4) = 4(x - 2)(x + 2)

But maybe easier to use substitution.

Let u = 4x² - 16 → du = 8x dx → x dx = du/8

Then:

∫ 9x / (4x² - 16) dx = 9 ∫ x dx / (4x² - 16) = 9 ∫ (du/8) / u = (9/8) ∫ du/u = (9/8) ln|u| + C

= (9/8) ln|4x² - 16| + C

We can simplify inside log:

|4x² - 16| = 4|x² - 4| → ln(4|x² - 4|) = ln 4 + ln|x² - 4|

But since we’re adding constant C, we can absorb ln 4 into it.

Alternatively, leave as is or write:

= (9/8) ln|4(x² - 4)| + C = (9/8)[ln 4 + ln|x² - 4|] + C = (9/8) ln|x² - 4| + C'

Either form is acceptable, but simplest is:

→ (9/8) ln|4x² - 16| + C

Or even better, factor out 4:

Since 4x² - 16 = 4(x² - 4), and ln|4(x² - 4)| = ln 4 + ln|x² - 4|, so:

Final simplified version: (9/8) ln|x² - 4| + C (since ln 4 is constant)

But let’s stick with original substitution result unless instructed otherwise.

Actually, let me double-check:

Original: ∫ 9x/(4x² - 16) dx

Factor numerator and denominator? Not necessary.

Substitution is solid.

Answer: (9/8) ln|4x² - 16| + C

But we can also write:

= (9/8) ln|4(x² - 4)| + C = (9/8)[ln 4 + ln|x² - 4|] + C = (9/8) ln|x² - 4| + K, where K = (9/8) ln 4 + C

Both are correct. For simplicity, I’ll present the factored version without the constant multiplier inside log.

Actually, in many textbooks, they prefer to factor constants out of logs when possible.

So better answer: (9/8) ln|x² - 4| + C

Wait — let’s verify by differentiating:

d/dx [ (9/8) ln|x² - 4| ] = (9/8) * (2x)/(x² - 4) = (18x)/(8(x² - 4)) = (9x)/(4(x² - 4)) = 9x/(4x² - 16) ✔ Correct.

So yes, (9/8) ln|x² - 4| + C is simpler and correct.

I’ll go with that.

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d) ∫ 1 / (16 + 4r²) dr

Factor denominator: 16 + 4r² = 4(4 + r²)

So:

∫ 1/[4(4 + r²)] dr = (1/4) ∫ 1/(r² + 4) dr

Now, ∫ 1/(r² + a²) dr = (1/a) arctan(r/a) + C

Here, a² = 4 → a = 2

So:

(1/4) * (1/2) arctan(r/2) + C = (1/8) arctan(r/2) + C

✔ Checked — correct.

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e) ∫ e^(3x) / (1 + e^(6x)) dx

Note: e^(6x) = (e^(3x))²

Let u = e^(3x) → du = 3e^(3x) dx → e^(3x) dx = du/3

Then integral becomes:

∫ [1 / (1 + u²)] * (du/3) = (1/3) ∫ 1/(1 + u²) du = (1/3) arctan(u) + C = (1/3) arctan(e^(3x)) + C

✔ Checked — correct.

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f) ∫ sin t / (1 + cos² t) dt

Let u = cos t → du = -sin t dt → sin t dt = -du

Integral becomes:

∫ (-du) / (1 + u²) = - ∫ 1/(1 + u²) du = - arctan(u) + C = - arctan(cos t) + C

✔ Checked — correct.

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Now, compiling all final answers:

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Final Answer:
a) (1/5) arctan(x/5) + C
b) (1/2) ln(x² + 4) + C
c) (9/8) ln|x² - 4| + C
d) (1/8) arctan(r/2) + C
e) (1/3) arctan(e^(3x)) + C
f) -arctan(cos t) + C