(i)  n x – + C (ii) – sin2x – cos2x + C

(iii) tan –1 x – + tan –1 x + C

(iv) x (  nx – 1) + C (v) +  n |sec x + tanx | + C

(vi) (x 2 – 1) + C (vii) x sin –1 + sin –1 + C

(viii) x tan –1 x –  n (1 + x 2 ) – + C

(ix) (sinx – cosx) + C (x) e x tanx + C

Sol. (i) =  nx. – =  nx. – + C

(ii) = dx = – dx

= – . + + C = – sin 2x – cos 2x + C

(iii)

= tan –1 x. – dx = tan –1 x – +

= tan –1 x – + tan –1 x + C

(iv) = = x lnx – = x  nx – x + C

(v) Let t = tanx

dt = sec 2 x dx

sec 2 x dx = =  n |t + | + C

= + ln |secx + tanx| + C

(vi) Ι = Put x 2 = t ⇒ 2x dx = dt

⇒ Ι = = t.e t – e t + C = – + C

(vii) Ι = put = t ⇒ dx = dt

⇒ Ι = 2 = 2 sin –1 t. – 2 dt = t 2 sin –1 t + –

= t 2 sin –1 t + + sin –1 t – sin –1 t + C = x sin –1 + – sin –1 + C

(viii) Let tan –1 x = t ⇒ dx = dt

⇒ = =

= t tant –  n |sect| –

= x tan –1 x –  n(1 + x 2 ) –

(ix) Ι =

= – e x cos x + = – e x cos x + e x sin x – dx

Ι = e x (sin x – cos x) + C

(x) = e x tanx + C

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