Question

Evaluate the integral using an appropriate trigonometric substitution dt V2 - 6t + 13 Evaluate the integral using an appropriate trigonometric substitution

Answer 1

dt dt V t-6t+13 ✓ +²-6t+9+4 dt V H-31+ substitute t-z=2 tanu. 15 2sec u dy dt= 2 secudu. 2 secu du (2 tanulty È sec²udy tantuti seci 11 /secudu. {/secudu= ln/secuttanul? 1) = lo secuttanulto

undo substitution tanu=t-3 secu = Sittantu Seca = √4+ (1-32 2 in Vates the 561415+ V63854_o =left-3+Jt-3) +91-402 +0. {lo ( 9 ) = lsa-lob} (c-loz) is also a constant. So we can replace (c-looz) as c. Elft3+ J€ 332+4)+c. v th-6t+13

2 JB x dx. / 2² / 6²-x2 dx. substitute x=bsinu s uzsin(x) dx=bcosudu. (256-27 dx = { b sinu 16 lbsinuj? (bcosuldu = / 6 sinu cosu bul-sinu du. = /64 sinžu costu du = 164 sin’ull-sin’u) du. = (64 sinhu- be sint uldu = 64 /sinkudu - 64 / sinte du Apply reduction formula. Isinnu du = 12-1/sinnadu- cosusin 26972"/sinºudu - cosu sinu 2 J -644- sinkudu - cosu sinu te

=6* [z lidu - cosusinu] - b4 [3 /sorudu - cesugiau) de + [ - cosusinu] ((2+ lidu - cosusion) - cosusin] = 6(4- cosusiny)-43 C - Cosusina) - cosusin) * =6* [*- cosusinu 3-6* [es - 3 casusing – cosusinky to agus b'cousiou - 3efu + 3.'cosusinu + b'cosusintu le cousiou o custou 4 ayda Du 8

undo substitution us sinu = 2 cosu= √1-sintu cosu V VB- 4 8 JE PJE, sirl:) 8 b'sin" (1) Now apply limits se deelper - Bovat + b*sin Y097 tb sin- 8

- [o-o+ b*splo]-[04-046] co