Miscellaneous Exercise 7: integration problems

This lesson works through six integrals from Miscellaneous Exercise 7. Each one uses a key idea: simplifying first, substitution, partial fractions, or a trigonometric substitution. The recurring tool is the identity $a^{\log m} = m$, which lets an exponential of a logarithm collapse to a plain power.

Question 8

Evaluate

$I = \int \frac{a^{5\log x} - a^{4\log x}}{a^{3\log x} - a^{2\log x}}\, dx.$

Using $m\log x = \log x^{m}$ and then $a^{\log m} = m$, each term simplifies, for example $a^{5\log x} = a^{\log x^{5}} = x^{5}$. So

$I = \int \frac{x^{5} - x^{4}}{x^{3} - x^{2}}\, dx.$

Factor the top as $x^{4}(x-1)$ and the bottom as $x^{2}(x-1)$. The $(x-1)$ cancels:

$I = \int \frac{x^{4}}{x^{2}}\, dx = \int x^{2}\, dx = \frac{x^{3}}{3} + c.$

Question 12

Evaluate

$I = \int \frac{x^{3}}{\sqrt{1 - x^{8}}}\, dx.$

Write $x^{8} = (x^{4})^{2}$, so the root is $\sqrt{1 - (x^{4})^{2}}$. Put $x^{4} = t$, so $4x^{3}\,dx = dt$, giving $x^{3}\,dx = \tfrac{1}{4}\,dt$. Then

$I = \frac{1}{4}\int \frac{dt}{\sqrt{1 - t^{2}}} = \frac{1}{4}\sin^{-1} t + c.$

Using the standard result $\int \dfrac{dt}{\sqrt{1 - t^{2}}} = \sin^{-1} t$ and resubstituting,

$I = \frac{1}{4}\sin^{-1}\!\left(x^{4}\right) + c.$

Question 13

Evaluate

$I = \int \frac{a^{x}}{(1 + a^{x})(2 + a^{x})}\, dx.$

Put $a^{x} = t$ so that $a^{x}\,dx = dt$. Then

$I = \int \frac{dt}{(1 + t)(2 + t)}.$

Partial fractions

Write $\dfrac{1}{(1+t)(2+t)} = \dfrac{A}{1+t} + \dfrac{B}{2+t}$, so $1 = A(2+t) + B(1+t)$.

• Put $t = -1$: $1 = A(1)$, so $A = 1$.
• Put $t = -2$: $1 = B(-1)$, so $B = -1$.

Therefore

$I = \int \frac{dt}{1+t} - \int \frac{dt}{2+t} = \log|1 + t| - \log|2 + t| + c.$

Resubstituting $t = a^{x}$ and using $\log a - \log b = \log\tfrac{a}{b}$,

$I = \log\left|\frac{1 + a^{x}}{2 + a^{x}}\right| + c.$

Question 14

Evaluate

$I = \int \frac{dx}{(x^{2} + 1)(x^{2} + 4)}.$

Put $x^{2} = t$. Because there is no $x$ term in the numerator, do not differentiate. Just do the partial fractions in $t$ and then rewrite $t$ back as $x^{2}$.

Partial fractions

Write $\dfrac{1}{(t+1)(t+4)} = \dfrac{A}{t+1} + \dfrac{B}{t+4}$, so $1 = A(t+4) + B(t+1)$.

• Put $t = -1$: $1 = 3A$, so $A = \tfrac{1}{3}$.
• Put $t = -4$: $1 = -3B$, so $B = -\tfrac{1}{3}$.

Rewriting $t = x^{2}$,

$I = \frac{1}{3}\int \frac{dx}{x^{2} + 1} - \frac{1}{3}\int \frac{dx}{x^{2} + 4}.$

Using $\int \dfrac{dx}{x^{2} + a^{2}} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$,

$I = \frac{1}{3}\tan^{-1} x - \frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2} + c = \frac{1}{3}\tan^{-1} x - \frac{1}{6}\tan^{-1}\frac{x}{2} + c.$

Question 15

Evaluate

$I = \int \cos^{3} x \cdot e^{\log \sin x}\, dx.$

Since $e^{\log \sin x} = \sin x$, this becomes

$I = \int \cos^{3} x \, \sin x \, dx.$

Put $\cos x = t$, so $-\sin x\,dx = dt$, that is $\sin x\,dx = -\,dt$. Then

$I = \int t^{3}(-\,dt) = -\frac{t^{4}}{4} + c = -\frac{\cos^{4} x}{4} + c.$

Question 19

Evaluate

$I = \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}\, dx.$

Put $\sqrt{x} = \cos t$, so $x = \cos^{2} t$ and $dx = -2\cos t \sin t \, dt$. Then

$I = \int \sqrt{\frac{1 - \cos t}{1 + \cos t}}\,(-2\cos t \sin t)\, dt.$

Use the half-angle identities $1 - \cos t = 2\sin^{2}\tfrac{t}{2}$, $1 + \cos t = 2\cos^{2}\tfrac{t}{2}$, and $\sin t = 2\sin\tfrac{t}{2}\cos\tfrac{t}{2}$. The square root becomes $\tan\tfrac{t}{2}$, and after simplifying,

$I = -4 \int \sin^{2}\frac{t}{2}\,\cos t \, dt.$

Replacing $\sin^{2}\tfrac{t}{2} = \tfrac{1 - \cos t}{2}$,

$I = -2 \int (1 - \cos t)\cos t \, dt = -2\int \cos t \, dt + 2\int \cos^{2} t \, dt.$

Using $\cos^{2} t = \tfrac{1 + \cos 2t}{2}$,

$I = -2\sin t + t + \frac{1}{2}\sin 2t + c.$

Back to $x$

Here $t = \cos^{-1}\sqrt{x}$. Using $\sin(\cos^{-1}\sqrt{x}) = \sqrt{1 - x}$, $\cos(\cos^{-1}\sqrt{x}) = \sqrt{x}$, and $\sin 2t = 2\sin t \cos t$,

$I = -2\sqrt{1 - x} + \cos^{-1}\sqrt{x} + \sqrt{1 - x}\,\sqrt{x} + c.$

Key takeaways

• $a^{\log m} = m$ and $e^{\log u} = u$, so an exponential of a logarithm collapses to a plain expression: always simplify before integrating.
• When you substitute $x^{2} = t$ and there is no $x$ in the numerator, do the partial fractions directly in $t$ and resubstitute; do not differentiate.
• A trigonometric substitution like $\sqrt{x} = \cos t$, together with the half-angle identities, turns an awkward square-root integral into simple cosine integrals.