Integrals containing a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 ugly because of the square root. WHY do we substitute trig functions? Because the Pythagorean identities turn a sum/difference of squares into a single squared term, so the root disappears cleanly.
1 − sin 2 θ = cos 2 θ ⇒ a 2 − a 2 sin 2 θ = a cos θ 1-\sin^2\theta=\cos^2\theta \Rightarrow \sqrt{a^2-a^2\sin^2\theta}=a\cos\theta 1 − sin 2 θ = cos 2 θ ⇒ a 2 − a 2 sin 2 θ = a cos θ 1 + tan 2 θ = sec 2 θ ⇒ a 2 + a 2 tan 2 θ = a sec θ 1+\tan^2\theta=\sec^2\theta \Rightarrow \sqrt{a^2+a^2\tan^2\theta}=a\sec\theta 1 + tan 2 θ = sec 2 θ ⇒ a 2 + a 2 tan 2 θ = a sec θ sec 2 θ − 1 = tan 2 θ ⇒ a 2 sec 2 θ − a 2 = a tan θ \sec^2\theta-1=\tan^2\theta \Rightarrow \sqrt{a^2\sec^2\theta-a^2}=a\tan\theta sec 2 θ − 1 = tan 2 θ ⇒ a 2 sec 2 θ − a 2 = a tan θ
We trade an algebraic root for a trig integral we already know how to do.
Definition Pattern → substitution table
Radical in integrand
Substitution
d x dx d x Root simplifies to
Range of θ \theta θ
a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 x = a sin θ x=a\sin\theta x = a sin θ d x = a cos θ d θ dx=a\cos\theta\,d\theta d x = a cos θ d θ a cos θ a\cos\theta a cos θ [ − π 2 , π 2 ] [-\tfrac\pi2,\tfrac\pi2] [ − 2 π , 2 π ]
a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x = a tan θ x=a\tan\theta x = a tan θ d x = a sec 2 θ d θ dx=a\sec^2\theta\,d\theta d x = a sec 2 θ d θ a sec θ a\sec\theta a sec θ ( − π 2 , π 2 ) (-\tfrac\pi2,\tfrac\pi2) ( − 2 π , 2 π )
x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x = a sec θ x=a\sec\theta x = a sec θ d x = a sec θ tan θ d θ dx=a\sec\theta\tan\theta\,d\theta d x = a sec θ tan θ d θ a tan θ a\tan\theta a tan θ [ 0 , π 2 ) ∪ [ π , 3 π 2 ) [0,\tfrac\pi2)\cup[\pi,\tfrac{3\pi}2) [ 0 , 2 π ) ∪ [ π , 2 3 π )
Here a > 0 a>0 a > 0 range restriction guarantees the root is non-negative so ⋅ = + \sqrt{\;\cdot\;}=+ ⋅ = +
Step 1 — substitute. Let x = a sin θ x=a\sin\theta x = a sin θ d x = a cos θ d θ dx=a\cos\theta\,d\theta d x = a cos θ d θ Why this step? The form a 2 − x 2 a^2-x^2 a 2 − x 2 1 − sin 2 θ 1-\sin^2\theta 1 − sin 2 θ only substitution that kills the root here.
Step 2 — simplify the root.
a 2 − x 2 = a 2 − a 2 sin 2 θ = a 2 cos 2 θ = a cos θ . \sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}=\sqrt{a^2\cos^2\theta}=a\cos\theta. a 2 − x 2 = a 2 − a 2 sin 2 θ = a 2 cos 2 θ = a cos θ . Why is it + a cos θ +a\cos\theta + a cos θ ± \pm ± Because θ ∈ [ − π 2 , π 2 ] \theta\in[-\tfrac\pi2,\tfrac\pi2] θ ∈ [ − 2 π , 2 π ] cos θ ≥ 0 \cos\theta\ge0 cos θ ≥ 0
Step 3 — assemble.
∫ a 2 − x 2 d x = ∫ a cos θ ⋅ a cos θ d θ = a 2 ∫ cos 2 θ d θ . \int\sqrt{a^2-x^2}\,dx=\int a\cos\theta\cdot a\cos\theta\,d\theta=a^2\int\cos^2\theta\,d\theta. ∫ a 2 − x 2 d x = ∫ a cos θ ⋅ a cos θ d θ = a 2 ∫ cos 2 θ d θ .
Step 4 — power-reduction. cos 2 θ = 1 + cos 2 θ 2 \cos^2\theta=\tfrac{1+\cos2\theta}{2} cos 2 θ = 2 1 + c o s 2 θ = a 2 2 ( θ + sin 2 θ 2 ) + C = a 2 2 θ + a 2 2 sin θ cos θ + C . =\frac{a^2}{2}\left(\theta+\frac{\sin2\theta}{2}\right)+C=\frac{a^2}{2}\theta+\frac{a^2}{2}\sin\theta\cos\theta+C. = 2 a 2 ( θ + 2 s i n 2 θ ) + C = 2 a 2 θ + 2 a 2 sin θ cos θ + C . Why this step? cos 2 \cos^2 cos 2
Step 5 — back-substitute using a right triangle: sin θ = x a ⇒ θ = arcsin x a \sin\theta=\frac{x}{a}\Rightarrow\theta=\arcsin\frac{x}{a} sin θ = a x ⇒ θ = arcsin a x cos θ = a 2 − x 2 a \cos\theta=\frac{\sqrt{a^2-x^2}}{a} cos θ = a a 2 − x 2 ∫ a 2 − x 2 d x = a 2 2 arcsin x a + x a 2 − x 2 2 + C \boxed{\int\sqrt{a^2-x^2}\,dx=\frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x\sqrt{a^2-x^2}}{2}+C} ∫ a 2 − x 2 d x = 2 a 2 arcsin a x + 2 x a 2 − x 2 + C
∫ d x 9 − x 2 \displaystyle\int\frac{dx}{\sqrt{9-x^2}} ∫ 9 − x 2 d x Step 1. a = 3 a=3 a = 3 x = 3 sin θ x=3\sin\theta x = 3 sin θ d x = 3 cos θ d θ dx=3\cos\theta\,d\theta d x = 3 cos θ d θ Why? Root is a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 Step 2. 9 − x 2 = 3 cos θ \sqrt{9-x^2}=3\cos\theta 9 − x 2 = 3 cos θ = ∫ 3 cos θ 3 cos θ d θ = ∫ d θ = θ + C . =\displaystyle\int\frac{3\cos\theta}{3\cos\theta}\,d\theta=\int d\theta=\theta+C. = ∫ 3 cos θ 3 cos θ d θ = ∫ d θ = θ + C . Why this step? The substitution made everything cancel — that's the payoff.
Step 3. θ = arcsin x 3 \theta=\arcsin\frac{x}{3} θ = arcsin 3 x = arcsin x 3 + C . =\arcsin\frac{x}{3}+C. = arcsin 3 x + C .
∫ d x x 2 x 2 + 4 \displaystyle\int\frac{dx}{x^2\sqrt{x^2+4}} ∫ x 2 x 2 + 4 d x Step 1. Root x 2 + 4 ⇒ a = 2 \sqrt{x^2+4}\Rightarrow a=2 x 2 + 4 ⇒ a = 2 x = 2 tan θ x=2\tan\theta x = 2 tan θ d x = 2 sec 2 θ d θ dx=2\sec^2\theta\,d\theta d x = 2 sec 2 θ d θ Step 2. x 2 + 4 = 2 sec θ \sqrt{x^2+4}=2\sec\theta x 2 + 4 = 2 sec θ x 2 = 4 tan 2 θ x^2=4\tan^2\theta x 2 = 4 tan 2 θ ∫ 2 sec 2 θ d θ 4 tan 2 θ ⋅ 2 sec θ = 1 4 ∫ sec θ tan 2 θ d θ . \int\frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot2\sec\theta}=\frac14\int\frac{\sec\theta}{\tan^2\theta}\,d\theta. ∫ 4 t a n 2 θ ⋅ 2 s e c θ 2 s e c 2 θ d θ = 4 1 ∫ t a n 2 θ s e c θ d θ . Step 3. Rewrite in sin/cos: sec θ tan 2 θ = 1 / cos θ sin 2 θ / cos 2 θ = cos θ sin 2 θ . \frac{\sec\theta}{\tan^2\theta}=\frac{1/\cos\theta}{\sin^2\theta/\cos^2\theta}=\frac{\cos\theta}{\sin^2\theta}. t a n 2 θ s e c θ = s i n 2 θ / c o s 2 θ 1/ c o s θ = s i n 2 θ c o s θ . Why? Now a clean u u u Step 4. Let u = sin θ u=\sin\theta u = sin θ 1 4 ∫ d u u 2 = − 1 4 u + C = − 1 4 sin θ + C . \frac14\int\frac{du}{u^2}=-\frac{1}{4u}+C=-\frac{1}{4\sin\theta}+C. 4 1 ∫ u 2 d u = − 4 u 1 + C = − 4 s i n θ 1 + C . Step 5. Triangle: tan θ = x 2 \tan\theta=\frac{x}{2} tan θ = 2 x = x =x = x = 2 =2 = 2 = x 2 + 4 =\sqrt{x^2+4} = x 2 + 4 sin θ = x x 2 + 4 \sin\theta=\frac{x}{\sqrt{x^2+4}} sin θ = x 2 + 4 x = − x 2 + 4 4 x + C . =-\frac{\sqrt{x^2+4}}{4x}+C. = − 4 x x 2 + 4 + C .
∫ x 2 − 1 x d x \displaystyle\int\frac{\sqrt{x^2-1}}{x}\,dx ∫ x x 2 − 1 d x Step 1. a = 1 a=1 a = 1 x = sec θ x=\sec\theta x = sec θ d x = sec θ tan θ d θ dx=\sec\theta\tan\theta\,d\theta d x = sec θ tan θ d θ Step 2. x 2 − 1 = tan θ \sqrt{x^2-1}=\tan\theta x 2 − 1 = tan θ = ∫ tan θ sec θ sec θ tan θ d θ = ∫ tan 2 θ d θ . =\displaystyle\int\frac{\tan\theta}{\sec\theta}\sec\theta\tan\theta\,d\theta=\int\tan^2\theta\,d\theta. = ∫ sec θ tan θ sec θ tan θ d θ = ∫ tan 2 θ d θ . Step 3. tan 2 θ = sec 2 θ − 1 \tan^2\theta=\sec^2\theta-1 tan 2 θ = sec 2 θ − 1 why? directly integrable): = tan θ − θ + C . =\tan\theta-\theta+C. = tan θ − θ + C . Step 4. Back: tan θ = x 2 − 1 \tan\theta=\sqrt{x^2-1} tan θ = x 2 − 1 θ = arcsec x = arccos 1 x \theta=\operatorname{arcsec}x=\arccos\frac1x θ = arcsec x = arccos x 1 = x 2 − 1 − arcsec x + C . =\sqrt{x^2-1}-\operatorname{arcsec}x+C. = x 2 − 1 − arcsec x + C .
Recall Predict the substitution BEFORE lookingFor each, name the sub, then check below.
∫ d x ( 4 − x 2 ) 3 / 2 \int\frac{dx}{(4-x^2)^{3/2}} ∫ ( 4 − x 2 ) 3/2 d x forecast: x = 2 sin θ x=2\sin\theta x = 2 sin θ ∫ x 2 x 2 − 25 d x \int\frac{x^2}{\sqrt{x^2-25}}dx ∫ x 2 − 25 x 2 d x forecast: x = 5 sec θ x=5\sec\theta x = 5 sec θ ∫ d x ( x 2 + 7 ) 2 \int\frac{dx}{(x^2+7)^2} ∫ ( x 2 + 7 ) 2 d x forecast: x = 7 tan θ x=\sqrt7\tan\theta x = 7 tan θ
All correct — the radical (or the ( ) 2 (\;)^2 ( ) 2
Common mistake Forgetting to convertd x dx d x Why it feels right: you mechanically replace x x x d x dx d x Fix: d x dx d x x = a tan θ ⇒ d x = a sec 2 θ d θ x=a\tan\theta\Rightarrow dx=a\sec^2\theta\,d\theta x = a tan θ ⇒ d x = a sec 2 θ d θ
a 2 cos 2 θ = a cos θ \sqrt{a^2\cos^2\theta}=a\cos\theta a 2 cos 2 θ = a cos θ Why it feels right: u 2 = u \sqrt{u^2}=u u 2 = u Fix: u 2 = ∣ u ∣ \sqrt{u^2}=|u| u 2 = ∣ u ∣ range restriction is precisely what makes cos θ ≥ 0 \cos\theta\ge0 cos θ ≥ 0 tan θ ≥ 0 \tan\theta\ge0 tan θ ≥ 0 ∣ ⋅ ∣ |\cdot| ∣ ⋅ ∣
Common mistake Leaving the answer inθ \theta θ Why it feels right: you "solved" the integral.
Fix: Build the reference right triangle from the substitution and re-express sin θ , cos θ , tan θ , θ \sin\theta,\cos\theta,\tan\theta,\theta sin θ , cos θ , tan θ , θ x x x
a sin θ a\sin\theta a sin θ x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 Why it feels right: sine is the "first" sub you learned.
Fix: Match the sign structure : minus-then-x 2 x^2 x 2 a 2 − x 2 a^2-x^2 a 2 − x 2 a 2 + x 2 a^2+x^2 a 2 + x 2 x 2 − a 2 x^2-a^2 x 2 − a 2
"SIN for the DIFFERENCE that starts with a a a x x x
Compact: a 2 − x 2 → sin a^2{-}x^2\to\sin a 2 − x 2 → sin a 2 + x 2 → tan a^2{+}x^2\to\tan a 2 + x 2 → tan x 2 − a 2 → sec x^2{-}a^2\to\sec x 2 − a 2 → sec S–T–S down the table, matching − , + , − -,+,- − , + , −
Recall Feynman: explain to a 12-year-oldImagine a ladder leaning on a wall. The ladder length, the wall height, and the floor distance always make a right triangle, and they're locked together by Pythagoras. A square root like a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 pretend x x x θ \theta θ . Once everything is angles, the square root vanishes (triangles always close up nicely), we solve the easy angle problem, then turn the angle back into the original side length using the triangle. We disguised a hard problem as a triangle, solved it, and undisguised it.
For a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 x = a sin θ x=a\sin\theta x = a sin θ ,
d x = a cos θ d θ dx=a\cos\theta\,d\theta d x = a cos θ d θ , root
→ a cos θ \to a\cos\theta → a cos θ .
For a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x = a tan θ x=a\tan\theta x = a tan θ ,
d x = a sec 2 θ d θ dx=a\sec^2\theta\,d\theta d x = a sec 2 θ d θ , root
→ a sec θ \to a\sec\theta → a sec θ .
For x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x = a sec θ x=a\sec\theta x = a sec θ ,
d x = a sec θ tan θ d θ dx=a\sec\theta\tan\theta\,d\theta d x = a sec θ tan θ d θ , root
→ a tan θ \to a\tan\theta → a tan θ .
Why does the root simplify under trig sub? Pythagorean identities convert sum/difference of squares into a single squared term.
Why must a 2 cos 2 θ = a cos θ \sqrt{a^2\cos^2\theta}=a\cos\theta a 2 cos 2 θ = a cos θ ± \pm ± The chosen
θ \theta θ -range keeps
cos θ ≥ 0 \cos\theta\ge0 cos θ ≥ 0 , so
∣ cos θ ∣ = cos θ |\cos\theta|=\cos\theta ∣ cos θ ∣ = cos θ .
∫ d x a 2 − x 2 = ? \int \frac{dx}{\sqrt{a^2-x^2}}=? ∫ a 2 − x 2 d x = ? arcsin x a + C \arcsin\frac{x}{a}+C arcsin a x + C .
∫ d x a 2 + x 2 = ? \int \frac{dx}{a^2+x^2}=? ∫ a 2 + x 2 d x = ? 1 a arctan x a + C \frac1a\arctan\frac{x}{a}+C a 1 arctan a x + C .
∫ a 2 − x 2 d x = ? \int\sqrt{a^2-x^2}\,dx=? ∫ a 2 − x 2 d x = ? a 2 2 arcsin x a + x a 2 − x 2 2 + C \frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x\sqrt{a^2-x^2}}{2}+C 2 a 2 arcsin a x + 2 x a 2 − x 2 + C .
Last mandatory step of any trig sub? Back-substitute to
x x x using the reference right triangle.
∫ tan 2 θ d θ = ? \int\tan^2\theta\,d\theta=? ∫ tan 2 θ d θ = ? tan θ − θ + C \tan\theta-\theta+C tan θ − θ + C (via
tan 2 = sec 2 − 1 \tan^2=\sec^2-1 tan 2 = sec 2 − 1 ).
Pythagorean identities — the engine that kills the root.Integration by substitution (u-sub) — trig sub is a clever change of variable.Power-reduction & double-angle formulas — needed for ∫ cos 2 θ , sin 2 θ \int\cos^2\theta,\sin^2\theta ∫ cos 2 θ , sin 2 θ Reference right triangle method — for back-substitution.Partial fractions — alternative for some rational integrals.Arc length and surface area — frequent source of 1 + x 2 \sqrt{1+x^2} 1 + x 2 Hyperbolic substitution — x = a sinh u x=a\sinh u x = a sinh u a cosh u a\cosh u a cosh u
Right triangle back-substitution
Closed-form antiderivative
Intuition Hinglish mein samjhoDekho, jab integral ke andar a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x x x 1 − sin 2 = cos 2 1-\sin^2=\cos^2 1 − sin 2 = cos 2 1 + tan 2 = sec 2 1+\tan^2=\sec^2 1 + tan 2 = sec 2 sec 2 − 1 = tan 2 \sec^2-1=\tan^2 sec 2 − 1 = tan 2
Rule simple hai: agar a 2 − x 2 a^2-x^2 a 2 − x 2 x = a sin θ x=a\sin\theta x = a sin θ a 2 + x 2 a^2+x^2 a 2 + x 2 x = a tan θ x=a\tan\theta x = a tan θ x 2 − a 2 x^2-a^2 x 2 − a 2 x = a sec θ x=a\sec\theta x = a sec θ d x dx d x d x dx d x
Last step sabse important: answer wapas x x x x x x θ \theta θ right triangle banao — substitution se ek side aur ek ratio mil jaati hai, baaki Pythagoras se nikal lo, aur phir sin θ \sin\theta sin θ cos θ \cos\theta cos θ θ \theta θ x x x
Common galti: a 2 cos 2 θ \sqrt{a^2\cos^2\theta} a 2 cos 2 θ a cos θ a\cos\theta a cos θ θ \theta θ cos θ \cos\theta cos θ
