← 2020 Paper 2
UPSC 2020 Maths Optional Paper 2 Q4b — Step-by-Step Solution 15 marks · Section A
Fundamental theorems of integral calculus · Real Analysis · asked 2× in 14 yrs · Read the full method →
Question
Show that ∫ 0 π / 2 sin 2 x sin x + cos x d x = 1 2 log e ( 1 + 2 ) \int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx=\frac{1}{\sqrt2}\log_e(1+\sqrt2) ∫ 0 π /2 s i n x + c o s x s i n 2 x d x = 2 1 log e ( 1 + 2 )
Technique
“King” reflection x → π 2 − x x\to\tfrac\pi2-x x → 2 π − x I = J I=J I = J 1 1 1 ∫ d x sin x + cos x \int\frac{dx}{\sin x+\cos x} ∫ s i n x + c o s x d x 2 sin ( x + π / 4 ) \sqrt2\sin(x+\pi/4) 2 sin ( x + π /4 ) ∫ csc = log ∣ tan ( t / 2 ) ∣ \int\csc=\log|\tan(t/2)| ∫ csc = log ∣ tan ( t /2 ) ∣ tan ( 3 π / 8 ) / tan ( π / 8 ) = ( 1 + 2 ) 2 \tan(3\pi/8)/\tan(\pi/8)=(1+\sqrt2)^2 tan ( 3 π /8 ) / tan ( π /8 ) = ( 1 + 2 ) 2
Solution
Let
I = ∫ 0 π / 2 sin 2 x sin x + cos x d x . I=\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx. I = ∫ 0 π /2 sin x + cos x sin 2 x d x . Step 1 — Use the King’s-rule symmetry x → π 2 − x x\to\frac{\pi}{2}-x x → 2 π − x
Apply the substitution x ↦ π 2 − x x\mapsto \tfrac{\pi}{2}-x x ↦ 2 π − x [ 0 , π 2 ] [0,\tfrac\pi2] [ 0 , 2 π ] sin ↔ cos \sin\leftrightarrow\cos sin ↔ cos
I = ∫ 0 π / 2 sin 2 ( π 2 − x ) sin ( π 2 − x ) + cos ( π 2 − x ) d x = ∫ 0 π / 2 cos 2 x cos x + sin x d x = : J . I=\int_0^{\pi/2}\frac{\sin^2(\tfrac\pi2-x)}{\sin(\tfrac\pi2-x)+\cos(\tfrac\pi2-x)}\,dx
=\int_0^{\pi/2}\frac{\cos^2 x}{\cos x+\sin x}\,dx=:J. I = ∫ 0 π /2 sin ( 2 π − x ) + cos ( 2 π − x ) sin 2 ( 2 π − x ) d x = ∫ 0 π /2 cos x + sin x cos 2 x d x =: J . So I = J I=J I = J
I + J = ∫ 0 π / 2 sin 2 x + cos 2 x sin x + cos x d x = ∫ 0 π / 2 d x sin x + cos x . I+J=\int_0^{\pi/2}\frac{\sin^2 x+\cos^2 x}{\sin x+\cos x}\,dx=\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}. I + J = ∫ 0 π /2 sin x + cos x sin 2 x + cos 2 x d x = ∫ 0 π /2 sin x + cos x d x . Since I = J I=J I = J
2 I = ∫ 0 π / 2 d x sin x + cos x ⟹ I = 1 2 ∫ 0 π / 2 d x sin x + cos x . ( ∗ ) 2I=\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}\quad\Longrightarrow\quad I=\frac12\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}.\tag{$\ast$} 2 I = ∫ 0 π /2 sin x + cos x d x ⟹ I = 2 1 ∫ 0 π /2 sin x + cos x d x . ( ∗ ) Step 3 — Evaluate K = ∫ 0 π / 2 d x sin x + cos x K=\displaystyle\int_0^{\pi/2}\frac{dx}{\sin x+\cos x} K = ∫ 0 π /2 sin x + cos x d x
Combine the denominator into a single sinusoid:
sin x + cos x = 2 sin ( x + π 4 ) . \sin x+\cos x=\sqrt2\,\sin\!\Big(x+\tfrac{\pi}{4}\Big). sin x + cos x = 2 sin ( x + 4 π ) . Hence
K = ∫ 0 π / 2 d x 2 sin ( x + π / 4 ) = 1 2 ∫ 0 π / 2 csc ( x + π 4 ) d x . K=\int_0^{\pi/2}\frac{dx}{\sqrt2\,\sin(x+\pi/4)}=\frac{1}{\sqrt2}\int_0^{\pi/2}\csc\!\Big(x+\tfrac{\pi}{4}\Big)\,dx. K = ∫ 0 π /2 2 sin ( x + π /4 ) d x = 2 1 ∫ 0 π /2 csc ( x + 4 π ) d x . Substitute t = x + π 4 t=x+\tfrac{\pi}{4} t = x + 4 π d t = d x dt=dx d t = d x x = 0 ⇒ t = π 4 x=0\Rightarrow t=\tfrac\pi4 x = 0 ⇒ t = 4 π x = π 2 ⇒ t = 3 π 4 x=\tfrac\pi2\Rightarrow t=\tfrac{3\pi}{4} x = 2 π ⇒ t = 4 3 π
K = 1 2 ∫ π / 4 3 π / 4 csc t d t . K=\frac{1}{\sqrt2}\int_{\pi/4}^{3\pi/4}\csc t\,dt. K = 2 1 ∫ π /4 3 π /4 csc t d t . Using ∫ csc t d t = log ∣ tan t 2 ∣ \displaystyle\int\csc t\,dt=\log\big|\tan\tfrac{t}{2}\big| ∫ csc t d t = log tan 2 t
K = 1 2 [ log ∣ tan t 2 ∣ ] π / 4 3 π / 4 = 1 2 ( log tan 3 π 8 − log tan π 8 ) = 1 2 log tan ( 3 π / 8 ) tan ( π / 8 ) . K=\frac{1}{\sqrt2}\Big[\log\big|\tan\tfrac{t}{2}\big|\Big]_{\pi/4}^{3\pi/4}
=\frac{1}{\sqrt2}\Big(\log\tan\tfrac{3\pi}{8}-\log\tan\tfrac{\pi}{8}\Big)
=\frac{1}{\sqrt2}\log\!\frac{\tan(3\pi/8)}{\tan(\pi/8)}. K = 2 1 [ log tan 2 t ] π /4 3 π /4 = 2 1 ( log tan 8 3 π − log tan 8 π ) = 2 1 log tan ( π /8 ) tan ( 3 π /8 ) . Step 4 — Simplify the half-angle tangents
Recall tan π 8 = 2 − 1 \tan\tfrac{\pi}{8}=\sqrt2-1 tan 8 π = 2 − 1 tan 3 π 8 = cot π 8 = 2 + 1 \tan\tfrac{3\pi}{8}=\cot\tfrac{\pi}{8}=\sqrt2+1 tan 8 3 π = cot 8 π = 2 + 1
tan ( 3 π / 8 ) tan ( π / 8 ) = 2 + 1 2 − 1 = ( 2 + 1 ) ( 2 − 1 ) ⋅ ( 2 + 1 ) ( 2 + 1 ) = ( 2 + 1 ) 2 1 = ( 2 + 1 ) 2 , \frac{\tan(3\pi/8)}{\tan(\pi/8)}=\frac{\sqrt2+1}{\sqrt2-1}=\frac{(\sqrt2+1)}{(\sqrt2-1)}\cdot\frac{(\sqrt2+1)}{(\sqrt2+1)}=\frac{(\sqrt2+1)^2}{1}=(\sqrt2+1)^2, tan ( π /8 ) tan ( 3 π /8 ) = 2 − 1 2 + 1 = ( 2 − 1 ) ( 2 + 1 ) ⋅ ( 2 + 1 ) ( 2 + 1 ) = 1 ( 2 + 1 ) 2 = ( 2 + 1 ) 2 , (using ( 2 − 1 ) ( 2 + 1 ) = 1 (\sqrt2-1)(\sqrt2+1)=1 ( 2 − 1 ) ( 2 + 1 ) = 1
K = 1 2 log ( ( 1 + 2 ) 2 ) = 1 2 ⋅ 2 log ( 1 + 2 ) = 2 2 log ( 1 + 2 ) = 2 log ( 1 + 2 ) . K=\frac{1}{\sqrt2}\log\big((1+\sqrt2)^2\big)=\frac{1}{\sqrt2}\cdot 2\log(1+\sqrt2)=\frac{2}{\sqrt2}\log(1+\sqrt2)=\sqrt2\,\log(1+\sqrt2). K = 2 1 log ( ( 1 + 2 ) 2 ) = 2 1 ⋅ 2 log ( 1 + 2 ) = 2 2 log ( 1 + 2 ) = 2 log ( 1 + 2 ) . Step 5 — Finish via ( ∗ ) (\ast) ( ∗ )
I = 1 2 K = 1 2 ⋅ 2 log ( 1 + 2 ) = 2 2 log ( 1 + 2 ) = 1 2 log ( 1 + 2 ) . I=\frac12 K=\frac12\cdot\sqrt2\,\log(1+\sqrt2)=\frac{\sqrt2}{2}\log(1+\sqrt2)=\frac{1}{\sqrt2}\log(1+\sqrt2). I = 2 1 K = 2 1 ⋅ 2 log ( 1 + 2 ) = 2 2 log ( 1 + 2 ) = 2 1 log ( 1 + 2 ) . Answer
∫ 0 π / 2 sin 2 x sin x + cos x d x = 1 2 log e ( 1 + 2 ) . \boxed{\;\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx=\frac{1}{\sqrt2}\,\log_e(1+\sqrt2).\;} ∫ 0 π /2 sin x + cos x sin 2 x d x = 2 1 log e ( 1 + 2 ) .