先求∫(0,x²)√(1+t²)dt和∫(x,2)t²xos(2t)dt的不定積分(∫(a,b)表示從a到b積分).
設(shè)t=tanα,則dt=sec²αdα,sinα=√[t/(1+t²)],cosα=1/√(1+t²)
∴不定積分∫√(1+t²)dt=∫sec³αdα
=∫d(sinα)/(1-sin²α)²
=(1/4)∫[1/(1+sinα)+1/(1+sinα)²+1/(1-sinα)+1/(1-sinα)²]d(sinα)
=(1/4)[ln(1+sinα)-1/(1+sinα)-ln(1-sinα)-1/(1-sinα)]+C (C是積分常數(shù))
=(1/4)[ln|(1+sinα)/(1-sinα)|-2/cos²α]+C
=(1/2)[ln|(1+sinα)/cosα|-1/cos²α]+C
=(1/2)[ln|√(1+t²)+√t|-t²-1]+C;
∴不定積分∫t²xos(2t)dt=(t²/2)sin(2t)-∫tsin(2t)dt (應(yīng)用分部積分)
=(t²/2)sin(2t)+(t/2)cos(2t)-(1/2)∫cos(2t)dt (應(yīng)用分部積分)
=(t²/2)sin(2t)+(t/2)cos(2t)-(1/4)sin(2t)+C (C是積分常數(shù))
故∫(0,x²)√(1+t²)dt=(1/2)[ln|√(1+t²)+√t|-t²-1]|(0,x²)
={ln[√(1+x^4)+x]-x^4}/2;
∫(x,2)t²xos(2t)dt=[(t²/2)sin(2t)+(t/2)cos(2t)-(1/4)sin(2t)]|(x,2)
=(7/4)sin4+cos4-(1/2)x²sin(2x)-(1/2)xcos(2x)+(2x)/4.