Integration by Substitution and Advanced Guessing
I Key FormulasKey Insight: Two different-looking antiderivatives are both right when their difference is a constant: 1 2 sin 2 x + c 1 = − 1 2 cos 2 x + c 2 \tfrac12\sin^2 x + c_1 = -\tfrac12\cos^2 x + c_2 2 1 sin 2 x + c 1 = − 2 1 cos 2 x + c 2 c 2 − c 1 = 1 2 c_2 - c_1 = \tfrac12 c 2 − c 1 = 2 1
II Guess, Differentiate, Fix the FactorA guess can be fixed exactly when its derivative is a constant multiple of the integrand.
III Choosing u u u Substitution works when the leftover factor becomes a constant multiple of d u du d u x x x
Integral
Example 1 · ∫ x³ (x⁴ + 2)⁵ dx
Example 2 · ∫ x dx / √(1 + x²)
Example 6 · ∫ dx / (x ln x)
∫ x 3 ( x 4 + 2 ) 5 d x \int x^3\,(x^4 + 2)^5\,dx ∫ x 3 ( x 4 + 2 ) 5 d x ∫ x d x 1 + x 2 \int \frac{x\,dx}{\sqrt{1 + x^2}} ∫ 1 + x 2 x d x ∫ d x x ln x , x > 0 \int \frac{dx}{x\ln x}, \qquad x > 0 ∫ x ln x d x , x > 0
Moves 1–2 · u and du
u = x 4 + 2 , d u = 4 x 3 d x u = x^4 + 2, \qquad du = 4x^3\,dx u = x 4 + 2 , d u = 4 x 3 d x Move 3 · rewrite in u, integrate
( x 4 + 2 ) 5 = u 5 , x 3 d x = 1 4 d u (x^4 + 2)^5 = u^5, \qquad x^3\,dx = \tfrac14\,du ( x 4 + 2 ) 5 = u 5 , x 3 d x = 4 1 d u ∫ x 3 ( x 4 + 2 ) 5 d x = 1 4 ∫ u 5 d u = u 6 4 ⋅ 6 + c = u 6 24 + c \int x^3 (x^4 + 2)^5\,dx = \tfrac14\int u^5\,du = \frac{u^6}{4 \cdot 6} + c = \frac{u^6}{24} + c ∫ x 3 ( x 4 + 2 ) 5 d x = 4 1 ∫ u 5 d u = 4 ⋅ 6 u 6 + c = 24 u 6 + c Move 4 · back to x
∫ x 3 ( x 4 + 2 ) 5 d x = 1 24 ( x 4 + 2 ) 6 + c \int x^3 (x^4 + 2)^5\,dx = \tfrac{1}{24}(x^4 + 2)^6 + c ∫ x 3 ( x 4 + 2 ) 5 d x = 24 1 ( x 4 + 2 ) 6 + c
Moves 1–2 · u and du
u = x 4 , d u = 4 x 3 d x u = x^4, \qquad du = 4x^3\,dx u = x 4 , d u = 4 x 3 d x Move 3 · rewrite in u, integrate
( x 4 + 2 ) 5 = ( u + 2 ) 5 , x 3 d x = 1 4 d u (x^4 + 2)^5 = (u + 2)^5, \qquad x^3\,dx = \tfrac14\,du ( x 4 + 2 ) 5 = ( u + 2 ) 5 , x 3 d x = 4 1 d u 1 4 ∫ ( u + 2 ) 5 d u = ( u + 2 ) 6 24 + c \tfrac14\int (u + 2)^5\,du = \frac{(u + 2)^6}{24} + c 4 1 ∫ ( u + 2 ) 5 d u = 24 ( u + 2 ) 6 + c Move 4 · back to x
∫ x 3 ( x 4 + 2 ) 5 d x = 1 24 ( x 4 + 2 ) 6 + c \int x^3 (x^4 + 2)^5\,dx = \tfrac{1}{24}(x^4 + 2)^6 + c ∫ x 3 ( x 4 + 2 ) 5 d x = 24 1 ( x 4 + 2 ) 6 + c
Moves 1–2 · u and du
u = x 3 , d u = 3 x 2 d x , d x = d u 3 x 2 u = x^3, \qquad du = 3x^2\,dx, \qquad dx = \frac{du}{3x^2} u = x 3 , d u = 3 x 2 d x , d x = 3 x 2 d u Move 3 · rewrite in u
x 3 d x = x 3 ⋅ d u 3 x 2 = x 3 d u x^3\,dx = x^3 \cdot \frac{du}{3x^2} = \frac{x}{3}\,du x 3 d x = x 3 ⋅ 3 x 2 d u = 3 x d u ∫ x 3 ( x 4 + 2 ) 5 d x = 1 3 ∫ x ( x 4 + 2 ) 5 d u \int x^3 (x^4 + 2)^5\,dx = \tfrac13\int x\,(x^4 + 2)^5\,du ∫ x 3 ( x 4 + 2 ) 5 d x = 3 1 ∫ x ( x 4 + 2 ) 5 d u The factor beside d u du d u x x x u u u
Moves 1–2 · u and du
u = 1 + x 2 , d u = 2 x d x u = 1 + x^2, \qquad du = 2x\,dx u = 1 + x 2 , d u = 2 x d x Move 3 · rewrite in u, integrate
x d x = 1 2 d u , 1 + x 2 = u 1 / 2 x\,dx = \tfrac12\,du, \qquad \sqrt{1 + x^2} = u^{1/2} x d x = 2 1 d u , 1 + x 2 = u 1/2 1 2 ∫ u − 1 / 2 d u = 1 2 ⋅ 2 u 1 / 2 + c = u 1 / 2 + c \tfrac12\int u^{-1/2}\,du = \tfrac12 \cdot 2u^{1/2} + c = u^{1/2} + c 2 1 ∫ u − 1/2 d u = 2 1 ⋅ 2 u 1/2 + c = u 1/2 + c Move 4 · back to x
∫ x d x 1 + x 2 = 1 + x 2 + c \int \frac{x\,dx}{\sqrt{1 + x^2}} = \sqrt{1 + x^2} + c ∫ 1 + x 2 x d x = 1 + x 2 + c
Moves 1–2 · u and du
u = 1 + x 2 , d u = 1 2 ( 1 + x 2 ) − 1 / 2 ⋅ 2 x d x = x d x 1 + x 2 u = \sqrt{1 + x^2}, \qquad du = \tfrac12(1 + x^2)^{-1/2} \cdot 2x\,dx = \frac{x\,dx}{\sqrt{1 + x^2}} u = 1 + x 2 , d u = 2 1 ( 1 + x 2 ) − 1/2 ⋅ 2 x d x = 1 + x 2 x d x Move 3 · rewrite in u, integrate
∫ x d x 1 + x 2 = ∫ d u = u + c \int \frac{x\,dx}{\sqrt{1 + x^2}} = \int du = u + c ∫ 1 + x 2 x d x = ∫ d u = u + c Move 4 · back to x
∫ x d x 1 + x 2 = 1 + x 2 + c \int \frac{x\,dx}{\sqrt{1 + x^2}} = \sqrt{1 + x^2} + c ∫ 1 + x 2 x d x = 1 + x 2 + c Here d u du d u
Moves 1–2 · u and du
u = x , d u = d x u = x, \qquad du = dx u = x , d u = d x Move 3 · rewrite in u
∫ x d x 1 + x 2 = ∫ u d u 1 + u 2 \int \frac{x\,dx}{\sqrt{1 + x^2}} = \int \frac{u\,du}{\sqrt{1 + u^2}} ∫ 1 + x 2 x d x = ∫ 1 + u 2 u d u The same integral with the letter renamed: u u u
Moves 1–2 · u and du
u = ln x , d u = d x x u = \ln x, \qquad du = \frac{dx}{x} u = ln x , d u = x d x Move 3 · rewrite in u, integrate
∫ 1 ln x ⋅ d x x = ∫ d u u = ln u + c \int \frac{1}{\ln x} \cdot \frac{dx}{x} = \int \frac{du}{u} = \ln u + c ∫ ln x 1 ⋅ x d x = ∫ u d u = ln u + c Move 4 · back to x
∫ d x x ln x = ln ( ln x ) + c \int \frac{dx}{x\ln x} = \ln(\ln x) + c ∫ x ln x d x = ln ( ln x ) + c For 0 < x < 1 0 < x < 1 0 < x < 1 ln x < 0 \ln x < 0 ln x < 0 ln ∣ ln x ∣ + c \ln\lvert\ln x\rvert + c ln ∣ ln x ∣ + c
Moves 1–2 · u and du
u = x ln x , d u = ( ln x + 1 ) d x u = x\ln x, \qquad du = (\ln x + 1)\,dx u = x ln x , d u = ( ln x + 1 ) d x Move 3 · rewrite in u
∫ d x x ln x = ∫ d u u ( ln x + 1 ) \int \frac{dx}{x\ln x} = \int \frac{du}{u\,(\ln x + 1)} ∫ x ln x d x = ∫ u ( ln x + 1 ) d u The factor 1 ln x + 1 \dfrac{1}{\ln x + 1} ln x + 1 1 x x x d u du d u
IV Two Answers, One Family1 2 sin 2 x \tfrac12\sin^2 x 2 1 sin 2 x − 1 2 cos 2 x -\tfrac12\cos^2 x − 2 1 cos 2 x 1 2 \tfrac12 2 1 x x x
V Quiz Questions
Problem 1 · Substitute the Inside
Given: ∫ x 2 ( x 3 + 1 ) 4 d x \displaystyle\int x^2\,(x^3 + 1)^4\,dx ∫ x 2 ( x 3 + 1 ) 4 d x u = x 3 + 1 u = x^3 + 1 u = x 3 + 1 Find the antiderivative.
a 1 5 ( x 3 + 1 ) 5 + c \tfrac15(x^3 + 1)^5 + c 5 1 ( x 3 + 1 ) 5 + c b 1 15 u 5 + c \tfrac{1}{15}u^5 + c 15 1 u 5 + c c 1 15 ( x 3 + 1 ) 5 + c \tfrac{1}{15}(x^3 + 1)^5 + c 15 1 ( x 3 + 1 ) 5 + c d 3 5 ( x 3 + 1 ) 5 + c \tfrac35(x^3 + 1)^5 + c 5 3 ( x 3 + 1 ) 5 + c
✅ Correct! x 2 d x = 1 3 d u x^2\,dx = \tfrac13\,du x 2 d x = 3 1 d u , so the integral is
1 3 ⋅ u 5 5 = 1 15 ( x 3 + 1 ) 5 + c \tfrac13 \cdot \tfrac{u^5}{5} = \tfrac{1}{15}(x^3+1)^5 + c 3 1 ⋅ 5 u 5 = 15 1 ( x 3 + 1 ) 5 + c .
❌ The factor 1 3 \tfrac13 3 1 d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x , so
x 2 d x x^2\,dx x 2 d x is only
1 3 d u \tfrac13\,du 3 1 d u . Differentiating
1 5 ( x 3 + 1 ) 5 \tfrac15(x^3+1)^5 5 1 ( x 3 + 1 ) 5 gives
3 x 2 ( x 3 + 1 ) 4 3x^2(x^3+1)^4 3 x 2 ( x 3 + 1 ) 4 , three times too big.
❌ Not yet the answer. The problem was posed in
x x x : change back with
u = x 3 + 1 u = x^3 + 1 u = x 3 + 1 .
❌ The 3 3 3 x 2 d x = 1 3 d u x^2\,dx = \tfrac13\,du x 2 d x = 3 1 d u , not
3 d u 3\,du 3 d u ; differentiating
3 5 ( x 3 + 1 ) 5 \tfrac35(x^3+1)^5 5 3 ( x 3 + 1 ) 5 gives
9 x 2 ( x 3 + 1 ) 4 9x^2(x^3+1)^4 9 x 2 ( x 3 + 1 ) 4 .
❌ Not quite. Write
d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x , replace
x 2 d x x^2\,dx x 2 d x by
1 3 d u \tfrac13\,du 3 1 d u , integrate
u 4 u^4 u 4 , then change back.
Show solution
Let u = x 3 + 1 u = x^3 + 1 u = x 3 + 1 d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x x 2 d x = 1 3 d u x^2\,dx = \tfrac13\,du x 2 d x = 3 1 d u
∫ x 2 ( x 3 + 1 ) 4 d x = 1 3 ∫ u 4 d u = u 5 3 ⋅ 5 + c = u 5 15 + c = 1 15 ( x 3 + 1 ) 5 + c . \int x^2 (x^3 + 1)^4\,dx = \tfrac13\int u^4\,du = \frac{u^5}{3 \cdot 5} + c = \frac{u^5}{15} + c = \tfrac{1}{15}(x^3 + 1)^5 + c. ∫ x 2 ( x 3 + 1 ) 4 d x = 3 1 ∫ u 4 d u = 3 ⋅ 5 u 5 + c = 15 u 5 + c = 15 1 ( x 3 + 1 ) 5 + c . Check: d d x [ 1 15 ( x 3 + 1 ) 5 ] = 5 15 ( x 3 + 1 ) 4 ⋅ 3 x 2 = x 2 ( x 3 + 1 ) 4 \dfrac{d}{dx}\Big[\tfrac{1}{15}(x^3 + 1)^5\Big] = \tfrac{5}{15}(x^3 + 1)^4 \cdot 3x^2 = x^2 (x^3 + 1)^4 d x d [ 15 1 ( x 3 + 1 ) 5 ] = 15 5 ( x 3 + 1 ) 4 ⋅ 3 x 2 = x 2 ( x 3 + 1 ) 4
Problem 2 · Write the Check, Signs Included
Given: ∫ x e − 3 x 2 d x \displaystyle\int x\,e^{-3x^2}\,dx ∫ x e − 3 x 2 d x Find the antiderivative by advanced guessing.
a 1 6 e − 3 x 2 + c \tfrac16 e^{-3x^2} + c 6 1 e − 3 x 2 + c b − 1 3 e − 3 x 2 + c -\tfrac13 e^{-3x^2} + c − 3 1 e − 3 x 2 + c c − 6 e − 3 x 2 + c -6\,e^{-3x^2} + c − 6 e − 3 x 2 + c d − 1 6 e − 3 x 2 + c -\tfrac16 e^{-3x^2} + c − 6 1 e − 3 x 2 + c
✅ Correct! The guess
e − 3 x 2 e^{-3x^2} e − 3 x 2 has derivative
− 6 x e − 3 x 2 -6x\,e^{-3x^2} − 6 x e − 3 x 2 , off by
− 6 -6 − 6 ; dividing by
− 6 -6 − 6 gives
− 1 6 e − 3 x 2 + c -\tfrac16 e^{-3x^2} + c − 6 1 e − 3 x 2 + c .
❌ The minus sign got lost. The derivative of the exponent
− 3 x 2 -3x^2 − 3 x 2 is
− 6 x -6x − 6 x , so the guess is off by
− 6 -6 − 6 , not
6 6 6 .
❌ The chain rule needs the derivative of − 3 x 2 -3x^2 − 3 x 2 − 6 x -6x − 6 x Differentiating
− 1 3 e − 3 x 2 -\tfrac13 e^{-3x^2} − 3 1 e − 3 x 2 gives
2 x e − 3 x 2 2x\,e^{-3x^2} 2 x e − 3 x 2 , twice the integrand.
❌ Multiplied instead of divided. The guess's derivative is
− 6 -6 − 6 times the integrand, so correct by
1 − 6 \tfrac{1}{-6} − 6 1 ;
− 6 e − 3 x 2 -6e^{-3x^2} − 6 e − 3 x 2 has derivative
36 x e − 3 x 2 36x\,e^{-3x^2} 36 x e − 3 x 2 .
❌ Not quite. Guess
e − 3 x 2 e^{-3x^2} e − 3 x 2 , write its derivative by the chain rule, and divide by the factor it is off by.
Show solution
Guess e − 3 x 2 e^{-3x^2} e − 3 x 2
d d x e − 3 x 2 = e − 3 x 2 ⋅ ( − 6 x ) = − 6 x e − 3 x 2 . \frac{d}{dx}e^{-3x^2} = e^{-3x^2} \cdot (-6x) = -6\,x\,e^{-3x^2}. d x d e − 3 x 2 = e − 3 x 2 ⋅ ( − 6 x ) = − 6 x e − 3 x 2 . That is − 6 -6 − 6 − 6 -6 − 6 k e − 3 x 2 k\,e^{-3x^2} k e − 3 x 2 − 6 k x e − 3 x 2 -6k\,x\,e^{-3x^2} − 6 k x e − 3 x 2 − 6 k = 1 -6k = 1 − 6 k = 1 k = − 1 6 k = -\tfrac16 k = − 6 1
∫ x e − 3 x 2 d x = − 1 6 e − 3 x 2 + c . \int x\,e^{-3x^2}\,dx = -\tfrac16 e^{-3x^2} + c. ∫ x e − 3 x 2 d x = − 6 1 e − 3 x 2 + c .
Problem 3 · The Integral of Tangent
Given: ∫ tan x d x = ∫ sin x cos x d x \displaystyle\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx ∫ tan x d x = ∫ cos x sin x d x u = cos x u = \cos x u = cos x Find the integral in u u u x x x
What does the integral become in u u u
a ∫ d u u \displaystyle\int \frac{du}{u} ∫ u d u b − ∫ u d u \displaystyle -\int u\,du − ∫ u d u c − ∫ d u u \displaystyle -\int \frac{du}{u} − ∫ u d u d ∫ sin x u d u \displaystyle \int \frac{\sin x}{u}\,du ∫ u sin x d u
What is ∫ tan x d x \int \tan x\,dx ∫ tan x d x
a − ln ∣ cos x ∣ + c -\ln\lvert\cos x\rvert + c − ln ∣ cos x ∣ + c b ln ∣ cos x ∣ + c \ln\lvert\cos x\rvert + c ln ∣ cos x ∣ + c c − ln ∣ sin x ∣ + c -\ln\lvert\sin x\rvert + c − ln ∣ sin x ∣ + c d − ln ∣ u ∣ + c -\ln\lvert u\rvert + c − ln ∣ u ∣ + c
✅ Correct! sin x d x = − d u \sin x\,dx = -du sin x d x = − d u , so
∫ tan x d x = − ∫ d u u = − ln ∣ cos x ∣ + c \int \tan x\,dx = -\int \frac{du}{u} = -\ln\lvert\cos x\rvert + c ∫ tan x d x = − ∫ u d u = − ln ∣ cos x ∣ + c .
❌ Check the sign of d u du d u d d x cos x = − sin x \frac{d}{dx}\cos x = -\sin x d x d cos x = − sin x , so
d u = − sin x d x du = -\sin x\,dx d u = − sin x d x and
sin x d x = − d u \sin x\,dx = -du sin x d x = − d u .
❌ u = cos x u = \cos x u = cos x 1 cos x \frac{1}{\cos x} c o s x 1 becomes
1 u \frac{1}{u} u 1 , not
u u u .
❌ An x x x The whole of
sin x d x \sin x\,dx sin x d x is
− d u -du − d u ; once it is replaced, no
sin x \sin x sin x remains.
❌ The sign from d u = − sin x d x du = -\sin x\,dx d u = − sin x d x Differentiating
ln ∣ cos x ∣ \ln\lvert\cos x\rvert ln ∣ cos x ∣ gives
− sin x cos x = − tan x \frac{-\sin x}{\cos x} = -\tan x c o s x − s i n x = − tan x .
❌ Change back with u = cos x u = \cos x u = cos x sin x \sin x sin x Differentiating
− ln ∣ sin x ∣ -\ln\lvert\sin x\rvert − ln ∣ sin x ∣ gives
− cot x -\cot x − cot x .
❌ Still in u u u Change back:
u = cos x u = \cos x u = cos x .
❌ Check the rewrite. Find
d u du d u for
u = cos x u = \cos x u = cos x , then replace
sin x d x \sin x\,dx sin x d x and
cos x \cos x cos x .
❌ Check the answer. Integrate
− 1 u -\frac{1}{u} − u 1 and change back to
x x x .
Show solution
With u = cos x u = \cos x u = cos x d u = − sin x d x du = -\sin x\,dx d u = − sin x d x sin x d x = − d u \sin x\,dx = -du sin x d x = − d u 1 cos x = 1 u \frac{1}{\cos x} = \frac{1}{u} c o s x 1 = u 1
∫ sin x cos x d x = − ∫ d u u = − ln ∣ u ∣ + c = − ln ∣ cos x ∣ + c . \int \frac{\sin x}{\cos x}\,dx = -\int \frac{du}{u} = -\ln\lvert u\rvert + c = -\ln\lvert\cos x\rvert + c. ∫ cos x sin x d x = − ∫ u d u = − ln ∣ u ∣ + c = − ln ∣ cos x ∣ + c . Check: d d x [ − ln ∣ cos x ∣ ] = − − sin x cos x = tan x \dfrac{d}{dx}\big[-\ln\lvert\cos x\rvert\big] = -\dfrac{-\sin x}{\cos x} = \tan x d x d [ − ln ∣ cos x ∣ ] = − cos x − sin x = tan x cos x ≠ 0 \cos x \ne 0 cos x = 0
Problem 4 · A Third Answer for sin x cos x \sin x \cos x sin x cos x
Given: Since sin x cos x = 1 2 sin 2 x \sin x\cos x = \tfrac12\sin 2x sin x cos x = 2 1 sin 2 x ∫ sin x cos x d x = − 1 4 cos 2 x + c 3 \displaystyle\int \sin x\cos x\,dx = -\tfrac14\cos 2x + c_3 ∫ sin x cos x d x = − 4 1 cos 2 x + c 3 Find c 3 − c 1 c_3 - c_1 c 3 − c 1 − 1 4 cos 2 x + c 3 = 1 2 sin 2 x + c 1 -\tfrac14\cos 2x + c_3 = \tfrac12\sin^2 x + c_1 − 4 1 cos 2 x + c 3 = 2 1 sin 2 x + c 1 x x x
a c 3 − c 1 = 1 2 c_3 - c_1 = \tfrac12 c 3 − c 1 = 2 1 b c 3 − c 1 = 1 4 c_3 - c_1 = \tfrac14 c 3 − c 1 = 4 1 c c 3 − c 1 = 0 c_3 - c_1 = 0 c 3 − c 1 = 0 d c 3 − c 1 = − 1 4 c_3 - c_1 = -\tfrac14 c 3 − c 1 = − 4 1
✅ Correct! c 3 − c 1 = 1 2 sin 2 x + 1 4 cos 2 x = 1 2 sin 2 x + 1 4 ( 1 − 2 sin 2 x ) = 1 4 c_3 - c_1 = \tfrac12\sin^2 x + \tfrac14\cos 2x = \tfrac12\sin^2 x + \tfrac14(1 - 2\sin^2 x) = \tfrac14 c 3 − c 1 = 2 1 sin 2 x + 4 1 cos 2 x = 2 1 sin 2 x + 4 1 ( 1 − 2 sin 2 x ) = 4 1 , a constant, so all three formulas name the same family.
❌ That is the shift between 1 2 sin 2 x \tfrac12\sin^2 x 2 1 sin 2 x − 1 2 cos 2 x -\tfrac12\cos^2 x − 2 1 cos 2 x A different pair of formulas has its own shift: compute
1 2 sin 2 x + 1 4 cos 2 x \tfrac12\sin^2 x + \tfrac14\cos 2x 2 1 sin 2 x + 4 1 cos 2 x with
cos 2 x = 1 − 2 sin 2 x \cos 2x = 1 - 2\sin^2 x cos 2 x = 1 − 2 sin 2 x .
❌ The formulas differ, so the constants must too. At
x = 0 x = 0 x = 0 :
− 1 4 + c 3 = 0 + c 1 -\tfrac14 + c_3 = 0 + c_1 − 4 1 + c 3 = 0 + c 1 , so
c 3 − c 1 ≠ 0 c_3 - c_1 \ne 0 c 3 − c 1 = 0 .
❌ That is c 1 − c 3 c_1 - c_3 c 1 − c 3 Collect the constants:
c 3 − c 1 = 1 2 sin 2 x + 1 4 cos 2 x c_3 - c_1 = \tfrac12\sin^2 x + \tfrac14\cos 2x c 3 − c 1 = 2 1 sin 2 x + 4 1 cos 2 x .
Show solution
Check the guess: d d x [ − 1 4 cos 2 x ] = 1 4 ⋅ 2 sin 2 x = 1 2 sin 2 x = sin x cos x \dfrac{d}{dx}\big[-\tfrac14\cos 2x\big] = \tfrac14 \cdot 2\sin 2x = \tfrac12\sin 2x = \sin x\cos x d x d [ − 4 1 cos 2 x ] = 4 1 ⋅ 2 sin 2 x = 2 1 sin 2 x = sin x cos x
Compare: set the two formulas equal and collect the constants on one side:
c 3 − c 1 = 1 2 sin 2 x + 1 4 cos 2 x = 1 2 sin 2 x + 1 4 ( 1 − 2 sin 2 x ) = 1 4 . c_3 - c_1 = \tfrac12\sin^2 x + \tfrac14\cos 2x = \tfrac12\sin^2 x + \tfrac14\big(1 - 2\sin^2 x\big) = \tfrac14. c 3 − c 1 = 2 1 sin 2 x + 4 1 cos 2 x = 2 1 sin 2 x + 4 1 ( 1 − 2 sin 2 x ) = 4 1 . The difference is the constant 1 4 \tfrac14 4 1 x x x x = 0 x = 0 x = 0 − 1 4 + c 3 = c 1 -\tfrac14 + c_3 = c_1 − 4 1 + c 3 = c 1 1 4 \tfrac14 4 1 c 1 c_1 c 1 c 2 − c 1 = 1 2 c_2 - c_1 = \tfrac12 c 2 − c 1 = 2 1 1 4 \tfrac14 4 1 c 2 c_2 c 2
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