MathematicsClass 12ICSE

How One Substitution Handles Integrals of Sine and Cosine Mixtures

Integrate reciprocals of linear combinations of sine and cosine and of their squares, integrate one sine-cosine combination divided by another, and use power substitutions for integrals such as 1 over x times x to the n plus 1.

Why do integrals mixing sine and cosine need special methods?

An integrand such as

\frac{1}{3 sin x + 4 cos x}

\frac{2 sin x + 3 cos x}{3 sin x + 4 cos x}

does not match any simple standard form. A few reliable techniques — rewriting as a single sine, dividing by

cos^{2} x

, or splitting the numerator — turn each one into something familiar.

This lesson covers reciprocals of sine-cosine combinations, ratios of such combinations, and power substitutions.

How do you integrate 1 over a linear combination of sine and cosine, or over a combination of their squares?

**For

\frac{1}{a sin x + b cos x}

, write the denominator as

r sin (x + α)

with

r = a^{2} + b^{2}

and integrate a cosecant; for

\frac{1}{a sin ^{2} x + b cos ^{2} x}

, divide above and below by

cos^{2} x

and substitute

t = tan x

.

Worked example (linear combination).** Find

\int \frac{d x}{3 sin x + cos x}

. Here

r = 2

and

3 sin x + cos x = 2 sin (x + \frac{π}{6})

, so

\int \frac{d x}{2 sin ( x + \frac{π}{6} )} = \frac{1}{2} lo g tan (\frac{x}{2} + \frac{π}{12}) + C

using

\int cosec u d u = lo g tan \frac{u}{2} + C

.

Worked example (squares). Find

\int \frac{d x}{4 sin ^{2} x + 9 cos ^{2} x}

. Dividing by

cos^{2} x

gives

\int \frac{sec ^{2} x d x}{4 tan ^{2} x + 9}

, and with

t = tan x

\int \frac{d t}{4 t ^{2} + 9} = \frac{1}{4} \times \frac{2}{3} tan^{- 1} \frac{2 t}{3} = \frac{1}{6} tan^{- 1} (\frac{2 tan x}{3}) + C

An everyday example. Two alternating currents slightly out of step add up to a single sine wave of the form

r sin (x + α)

— the same combination used in the first example.

The substance. **The substitution

t = tan \frac{x}{2}

always works** — it turns any rational function of sine and cosine into a rational function of t, though the shortcuts above are usually quicker.

How do you integrate one linear combination of sine and cosine divided by another?

**For

\int \frac{a sin x + b cos x}{c sin x + d cos x} d x

, write the numerator as A times the denominator plus B times the derivative of the denominator, find A and B by comparing coefficients, and the integral is

A x + B lo g ∣ c sin x + d cos x ∣ + C

.

Worked example.** Find

\int \frac{2 sin x + 3 cos x}{3 sin x + 4 cos x} d x

.

- The denominator is

3 sin x + 4 cos x

, with derivative

3 cos x - 4 sin x

- Write

2 sin x + 3 cos x = A (3 sin x + 4 cos x) + B (3 cos x - 4 sin x)

- Sine terms:

3 A - 4 B = 2

; cosine terms:

4 A + 3 B = 3

- Solving gives

A = \frac{18}{25}

and

B = \frac{1}{25}

\int \frac{2 sin x + 3 cos x}{3 sin x + 4 cos x} d x = \frac{18}{25} x + \frac{1}{25} lo g ∣3 sin x + 4 cos x ∣ + C

Worked example 2.

\int \frac{d x}{1 + tan x} = \int \frac{cos x}{sin x + cos x} d x = \frac{x}{2} + \frac{1}{2} lo g ∣ sin x + cos x ∣ + C

.

An everyday example. Splitting a shared electricity bill into a fixed share and a share based on use mirrors splitting the numerator into a multiple of the denominator and a multiple of its derivative.

The substance. Two equations always fix A and B — a numerator made of sine and cosine has exactly two coefficients to match.

How do you evaluate integrals like 1 over x times x to the n plus 1 using substitution?

**For

\int \frac{d x}{x ( x ^{n} + 1 )}

, multiply above and below by

x^{n - 1}

and substitute

t = x^{n}

, so that

d t = n x^{n - 1} d x

and the integral becomes

\frac{1}{n} \int \frac{d t}{t ( t + 1 )}

, which partial fractions finish.

The general result:**

\int \frac{d x}{x ( x ^{n} + 1 )} = \frac{1}{n} lo g \frac{x ^{n}}{x ^{n} + 1} + C

Worked example. Find

\int \frac{d x}{x ( x ^{4} + 1 )}

.

- Multiply above and below by

x^{3}

\int \frac{x ^{3} d x}{x ^{4} ( x ^{4} + 1 )}

- Put

t = x^{4}

, so

d t = 4 x^{3} d x

\frac{1}{4} \int (\frac{1}{t} - \frac{1}{t + 1}) d t

- Result:

\frac{1}{4} lo g \frac{x ^{4}}{x ^{4} + 1} + C

Worked example 2. For

\int \frac{d x}{x ( x ^{3} - 1 )}

, the same steps with

t = x^{3}

give

\frac{1}{3} lo g \frac{x ^{3} - 1}{x ^{3}} + C

.

An everyday example. Grouping loose coins into stacks of ten before counting changes the unit to make the count easy, just as

t = x^{n}

changes the variable.

The substance. **The trick is to create the derivative of

x^{n}

** — multiplying by

x^{n - 1}

supplies exactly the factor the substitution needs.

Exam tip

What earns full marks on integrals of trigonometric combinations?

Show the values of A and B, or of r and the angle, on separate lines before integrating — they carry the method marks.

-

a sin x + b cos x = r sin (x + α)

with

r = a^{2} + b^{2}

- Squares of sine and cosine: divide by

cos^{2} x

and put

t = tan x

- Ratio of combinations: A times the denominator plus B times its derivative
-

\frac{1}{x ( x ^{n} + 1 )}

: multiply by

x^{n - 1}

and put

t = x^{n}

The trap. Writing

\int \frac{d x}{sin x + cos x}

lo g ∣ sin x + cos x ∣

. The numerator is not the derivative of the denominator, so the log form does not apply.

Did you know

Why can two sound waves cancel each other out?

A combination

a sin x + b cos x

is always a single wave,

r sin (x + α)

, with a new height r and a shifted starting point. Adding two waves of the same frequency never creates a strange shape — only a larger, smaller or shifted wave.

When two equal waves are exactly half a cycle apart, r becomes zero and they cancel completely. Noise-cancelling headphones use this idea, producing a wave that is the mirror image of the incoming noise.

The same identity that simplifies these integrals explains why the sound disappears.

Exam relevance

How are integrals of trigonometric combinations tested in JEE Main?

Integrals is a recurring JEE Main chapter, and these forms often reappear inside definite integrals with symmetric limits.

What gets asked. Ratios of sine-cosine combinations such as

\frac{1}{1 + tan x}

, **reciprocals of

a sin x + b cos x

**, integrals with

sin^{2} x

and

cos^{2} x

in the denominator, and power substitutions such as

\frac{1}{x ( x ^{n} + 1 )}

.

Question types. Multiple-choice questions asking for constants in a given answer form, and numerical-value questions.

The trap that costs marks. Getting the sign of the derivative wrong while finding A and B — the derivative of

cos x

- sin x

Key takeaways

What must you be able to do from this lesson?

- Reciprocal forms: rewrite

a sin x + b cos x

r sin (x + α)

; for squares, divide by

cos^{2} x

and put

t = tan x

- Ratios of combinations: numerator equals A times the denominator plus B times its derivative, giving

A x + B lo g ∣ denominator ∣

- Power substitutions:

\int \frac{d x}{x ( x ^{n} + 1 )} = \frac{1}{n} lo g \frac{x ^{n}}{x ^{n} + 1} + C

Can you find

\int \frac{d x}{1 + cot x}

Why do integrals mixing sine and cosine need special methods?

How do you integrate 1 over a linear combination of sine and cosine, or over a combination of their squares?

How do you integrate one linear combination of sine and cosine divided by another?

How do you evaluate integrals like 1 over x times x to the n plus 1 using substitution?

What earns full marks on integrals of trigonometric combinations?

Why can two sound waves cancel each other out?

How are integrals of trigonometric combinations tested in JEE Main?

What must you be able to do from this lesson?

Related study notes

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