Integrate the functions in Exercises 1 to 22.
Question 1: x sinx.
Solution:
Let f(X) = ∫ x sinx dx
Taking x as first function and sin x as second function and integrating by parts, we obtain
f(x) = x ∫sin x dx – {(d(x)/dx) ∫sin x dx} dx
⇒ f(x) = x(-cosx)-∫1. (-cosx)dx
⇒ f(x) = x cosx + sinx + C
Question 2: x sin3x
Solution:
Let f(x) = ∫x sin3x dx
Taking x as the first function and sin 3x as the second function and integrating by parts, we obtain
f(x) = x ∫sin3x dx – {(d(x)/dx) ∫sin3x dx} dx
⇒ f(x) = x (-cos3x ∕ 3) – ∫1. (cos3x ∕ 3) dx
⇒ f(x) = -x cos3x∕ 3 + 1∕3 ∫cos3x dx
⇒ f(x) = -xcos3x∕3 + 1∕9 sin3x + C
Question 3: x2 ex
Solution:
Let f(x) = ∫x2 ex dx
Taking x2 as first function and ex as second function and integrating by parts, we obtain
f(x) = x2 ∫ex dx – {(d(x2)/dx) ∫ex dx} dx
⇒ f(x) = x2 ex– ∫2x.ex dx
⇒ f(x) = x2ex – 2 ∫x.ex dx
Again, integrating by parts, we obtain
f(x) = x2ex -2[x.∫exdx-∫{(d(x)/dx). ∫exdx}dx]
⇒ f(x) = x2ex -2[xex – ∫exdx]
⇒ f(x) = x2ex-2[xex – ex]
⇒ f(x) = x2ex -2xex + 2ex + C
⇒ f(x) = ex (x2-2x+2) + C
Question 4: x logx
Solution:
Let f(x) = ∫x log x dx
Taking log x as first function and x as second function and integrating by parts, we obtain
f(x) = log x ∫x dx – {(d(log x)/dx) ∫x dx} dx
⇒ f(x) = log x. (x2 ∕ 2) – ∫1∕x. (x2 ∕2) dx
⇒ f(x) = x2logx∕2 – ∫x∕2 dx
⇒ f(x) = x2logx∕2 – x2∕4 + C
Question 5: x log2x
Solution:
Let f(x) = ∫x log 2x dx
Taking log 2x as first function and x as second function and integrating by parts, we obtain
f(x) = log 2x ∫x dx – {(d (log 2x)∕dx) ∫x dx} dx
⇒ f(x) = log 2x. (x2 ∕ 2) – ∫2∕2x. (x2 ∕2) dx
⇒ f(x) = x2log2x∕2 – ∫x∕2 dx
⇒ f(x) = x2log2x∕2 – x2∕4 + C
Question 6: x2 logx
Solution:
Let f(x) = ∫x2 log x dx
Taking log x as first function and x2 as second function and integrating by parts, we obtain
f(x) = log x ∫x2 dx – {(d (log x) ∕dx) ∫x2 dx} dx
⇒ f(x) = log x. (x3 ∕ 3) – ∫1∕x. (x3 ∕3) dx
⇒ f(x) = x3logx∕3 – ∫x3∕ 3 dx
⇒ f(x) = x3logx∕3 – x3 ∕ 9 + C
Question 7: x sin-1x.
Solution:
Let f(x) = ∫x sin-1x dx
Taking sin-1x as first function and x as second function and integrating by parts, we obtain
f(x) = sin-1x ∫ x dx – ∫ {(d(sin-1x∕dx) ∫x dx} dx
⇒ f(x) = sin-1x (x2/2) – ∫ 1∕ √(1-x2 ).x2∕2 dx
⇒ f(x) = x2sin-1x ∕2 + 1∕2 ∫-x ∕ √(1-x2 ) dx
⇒ f(x) = x2 sin-1x∕2 + 1∕2 ∫ {1-x2 ∕ √(1-x2 )- 1∕√(1-x2)} dx
⇒ f(x) = x2 sin-1x∕2 + 1∕2 ∫ {√1-x2 – 1∕√(1-x2)} dx
⇒ f(x) = x2 sin-1x∕2 + 1∕2 {∫ √1-x2 dx – 1 ∕√(1-x2) dx}
⇒ f(x) = x2sin-1x∕2 + 1∕2{x∕2. √1-x2 + 1∕2sin-1x – sin-1x} +C
⇒ f(x) = x2sin-1x∕2 + x∕4. √1-x2 + 1∕4sin-1x – 1∕2sin-1x +C
⇒ f(x) = 1∕4(2x2-1) sin-1x + x∕4. √1-x2 + C
Question 8: x tan-1x
Solution:
Let f(x) = ∫ x tan-1x dx
Taking tan-1x as first function and x as second function and integrating by parts, we obtain
f(x) = tan-1x ∫ x dx – ∫ {(d(tan-1x∕dx) ∫x dx} dx
⇒ f(x) = tan-1x (x2∕2) – ∫ 1∕ (1+x). x2∕2 dx
⇒ f(x) = x2 tan-1x ∕ 2 – 1∕2∫ 1∕ (1+x2 ) dx
⇒ f(x) = x2 tan-1x ∕ 2 – 1∕2∫{(x2+1) ∕ (1+x2) – 1∕ (1+x2)} dx
⇒ f(x) = x2 tan-1x ∕ 2 – 1∕2∫ (1- 1∕ (1+x2) dx
⇒ f(x) = x2 tan-1x ∕ 2 – 1∕2 (x-tan-1 x) + C
⇒ f(x) = x2 tan-1x ∕ 2 – x∕2 + 1∕2 tan-1x + C
Question 9: x cos-1x
Solution:
Let f(x) = ∫x cos-1x dx
Taking cos-1x as first function and x as second function and integrating by parts, we obtain
f(x) = cos-1x ∫ x dx – ∫ {(d(cos-1x)∕dx.∫x dx} dx
⇒ f(x) = cos-1x (x2/2) – ∫ -1∕ √1-x2 .x2∕2 dx
⇒ f(x) = x2 cos-1x ∕2 – 1∕2 ∫1-x2-1 ∕ √(1-x2 ) dx
⇒ f(x) = x2 cos-1x∕2 – 1∕2 ∫ {√(1-x2 ) +( -1∕√1-x2)} dx
⇒ f(x) = x2 cos-1x∕2 – 1∕2 ∫ √1-x2 dx – 1∕2 ∫(-1∕√1-x2) dx
⇒ f(x) = x2 cos-1x∕2 – 1∕2 I1 – 1∕2 cos-1x ————-(1)
Where I1 = ∫√1-x2 dx
I1 = x√1-x2 – ∫d(√1-x2)∕dx ∫x dx
⇒ I1 = x√1-x2 – ∫d(-2x∕2√1-x2 .x dx
⇒ I1 = x√1-x2 – ∫-x2∕√1-x2 dx
⇒ I1 = x√1-x2 – ∫1-x2-1 ∕ √(1-x2) dx
⇒ I1 = x√1-x2 – { ∫ √1-x2 dx + ∫(-dx∕√1-x2}
⇒ I1 = x√1-x2– {I1 + cos-1x}
⇒ 2I1 = x√1-x2 – cos-1x
⇒ I1 = x∕2 .√1-x2 -1∕2 cos-1x
Substituting in (1), we obtain
f(x) = x2cos-1x∕2 – 1∕2 (x∕2 .√1-x2 -1∕2 cos-1x) – 1∕2 cos-1x
⇒ f(x) = (2x-1)∕4 cos-1x – x∕4 √1-x2 + C
Question 10: (sin-1x)2
Solution:
Let f(x) = ∫(sin-1x)2 dx
Taking (sin-1x)2as first function and 1 as second function and integrating by parts, we obtain
f(x) = (sin-1x)2∫ 1dx – ∫ {(d(sin-1x)2∕dx. ∫1 dx} dx
⇒ f(x) = (sin-1x). x – ∫ 2.sin-1x∕√1-x2 . x dx
⇒ f(x) = x(sin-1x) 2 + ∫sin-1x.(-2x∕ √1-x2)dx
⇒ f(x) = x(sin-1x) 2+[sin-1x∫-2x∕√1-x2 dx – ∫ {❴d(sin-1x)∕dx❵∫-2x ∕√1-x2 dx}dx]
⇒ f(x) =x(sin-1x) 2+[ sin-1x .2√1-x2 – ∫1∕√1-x2 .2√1-x2 dx ]
⇒ f(x) =x(sin-1x)2+ 2√1-x2 sin-1x – ∫2dx
⇒ f(x) =x(sin-1x)2+ 2√1-x2 .sin-1x -2x + C
Question 11: (x cos-1x) / √1-x2
Solution:
Let f(x) = ∫(x cos-1x) / √1-x2 dx
we are multiplying -1/2 in numerator and dementor then.
f(x) = -1∕2 ∫(-2x cos-1x) ∕√1-x2 dx
Taking (cos-1x) as first function and {-2x∕√1-x2 ❵ as second function and integrating by parts, we obtain
f(x) = -1∕2 [ cos-1x∫-2x∕√1-x dx – ∫ {❲d(cos-1x)∕dx❳∫-2x√1-x2 dx}dx]
⇒ f(x) = -1∕2 [cos-1x.2√1-x2 – ∫-1∕√1-x2 2√1-x2 dx]
⇒ f(x) = -1∕2 [2√1-x2 cos-1x+ ∫2 dx]
⇒ f(x) = -1∕2 [2√1-x2 cos-1x+ 2x] + C
⇒ f(x) = – [√1-x2 cos-1x+ x] + C
Question 12: x sec2x
Solution:
Let f(x) = ∫x sec2x dx
Taking x as first function and sec2x as second function and integrating by parts, we obtain
f(x) = x ∫sec2x dx – {(d (x) ∕dx) ∫sec2x dx} dx
⇒ f(x) = xtanx – ∫1.tanx dx
⇒ f(x) = x tanx + log|cosx| + C
Question 13: tan-1x
Solution:
Let f(x) = ∫tan-1x dx
Taking tan-1x as first function and 1 as second function and integrating by parts, we obtain
f(x) = tan-1x∫1dx – ∫ {[d❲tan-1x❳∕dx] ∫1. dx}dx
⇒ f(x) = tan-1x .x – ∫1∕1+x2 .x dx
⇒ f(x) = x tan-1x – 1∕2 ∫2x∕1+x2 dx
⇒ f(x) = x tan-1x – 1∕2 log|1+x2| +C
⇒ f(x) = x tan-1x -1∕2 log(1+x2) + C
Question 14: x (logx)2
Solution:
Let f(x) = ∫ x (logx)2 dx
Taking (logx)2 as first function and x as second function and integrating by parts, we obtain
f(x) = (logx)2 ∫x dx – ∫[{{d(logx)∕dx}2}∫xdx]dx
⇒ f(x) = x∕2 (logx)2 – [∫2logx 1∕x . x2∕2 dx]
⇒ f(x) = x2∕2 (logx)2 – ∫x logx dx
Again, integration by parts, we obtain.
f(x) = x2∕2 ❲logx❳2 – [logx ∫x dx – ∫{❴d(logx)∕dx}∫xdx❵dx]
⇒ f(x) = x2∕2 ❲logx❳2 – [x2∕2 – logx – ∫1∕x .x2∕2 dx]
⇒ f(x) = x2∕2 ❲logx❳2 – x2∕2 .logx + 1∕2∫xdx
⇒ f(x) = x2∕2 ❲logx❳2 – x2∕2 .logx + x2∕4 + C
Question 15: (x2+1) logx
Solution:
Let f(x) = ∫ (x2+1) logx dx
⇒ f(x) = ∫ x2 logx dx + ∫ logx dx
Let f(x) = I1 + I2 ……………………….. (1)
where I1 = ∫ x2 logx dx and I2 = ∫ logx dx
I1 = ∫ x2 logx dx
Taking (logx) as first function and x2 as second function and integrating by parts, we obtain
⇒ I1 = logx – ∫ x2dx – ∫{❴d(logx)∕dx❵∫x2dx} dx
⇒ I1 = logx .x3∕3 – ∫1∕x . x3∕3 dx
⇒ I1 = x3∕3 logx – 1∕3(∫x dx)
⇒ I1 = x3∕3 logx – x3∕9 + C1 …………………… (2)
I2 = ∫ logx dx
Taking log x as first function and 1 as second function and integrating by parts, we obtain
f(x) = log x ∫1 dx – {(d(log x)/dx) ∫1 dx} dx
⇒ f(x)= log x.x – ∫1∕x. x dx
⇒ f(x) = x. logx∕2 – ∫1 dx
⇒ f(x) = x. logx∕2 – x + C2 ………………….(3)
Using equation (2) and (3) in (1), we obtain
f(x) = x3∕3 logx – x3∕9 + C1 + x. logx∕2 – x + C2
⇒ f(x)= x3∕3 logx – x3∕9 + x. logx∕2 – x + C 1+ C2
⇒ f(x)= (x3∕3 + x) logx – x3∕9 – x + C
Question 16: ex(sinx + cosx)
Solution:
Let f(x) = ∫ e(sinx+cosx) dx
Let g(x) = sinx
g‘(x) = cosx
f(x) = ∫ ex{g(x) + g‘(x) } dx
It is known that, ∫ ex{ g(x) + g'(x) } dx = ex g(x) + C
So, f(x) = ex sinx + C
Question 17: x ex ∕(1+x)2
Solution:
Let f(x) = ∫ x ex ∕(1+x)2 dx
⇒ f(x)= ∫ ex{x ∕(1+x)2} dx
⇒ f(x)= ∫ ex{(1+x-1) ∕ (1+x)2} dx
⇒ f(x)= ∫ e {1∕ (1+x) – 1∕ (1+x)2} dx
Let f(x) = 1∕ (1+x)
f‘(x) = -1∕ (1+x)2
⇒ f(x) = ∫ x ex ∕(1+x)2 dx = ∫ ex{f(x) + f'(x) }dx
It is known that ∫ ex{f(x) + f'(x) }dx = ex f(x) + C
So, ∫ x ex ∕(1+x)2 dx = ex ∕ (1+x) C
Question 18: ex{(1+sinx) ∕ (1+cosx)}
Solution:
Let I = ex{(1+sinx)∕(1+cosx)}
⇒ I =ex{(sin2x∕2+cos2x∕2+2 sinx∕2 cosx∕2) ∕ (2cos2x∕2)}
⇒ I = ex{(sinx∕2+cosx∕2)2 ∕ (2cos2x∕2)}
⇒ I = 1∕2 ex{(sinx∕2+cosx∕2) ∕ (cosx∕2)}2
⇒ I = 1∕2 ex {tanx∕2 + 1}2
⇒ I = 1∕2 ex{1 + tan2x∕2 + 2tanx∕2}
⇒ I = 1∕2 ex{secx∕2 + 2tanx∕2}
⇒ I = ex(1+sinx)dx ∕ (1+cosx) = ex{1∕2 sec2x∕2 + tanx∕2} —————– (1)
Let tanx∕2 = f(x) or f'(x) = 1∕2 sec2x∕2
It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C
from equation (1), we obtain,
∫ex{(1+sinx)∕(1+cosx)}dx = ex tanx∕2 + C
Question 19: ex{1∕x – 1∕x2}
Solution:
Let f(x) = ∫ex{1∕x – 1∕x2} dx
Also Let 1∕x = f(x) or f'(x) = – 1∕x2
It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C
So, f(x) = ex ∕ x + C
Question 20: (x-3) ex ∕ (x-1)3
Solution:
Let f(x) = ∫ ex{ (x-3) ∕ (x-1)3}dx
⇒ f(x) = ∫ ex{ (x-1-2) ∕ (x-1)3}dx
⇒ f(x) = ∫ ex{ 1∕ (x-1)2 – 2 ∕ (x-1)3}dx
⇒ f(x) = 1 ∕ (x-1)2 or f'(x) = -2 ∕ (x-1)3
It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C
So, ∫ ex{ (x-3) ∕ (x-1)3}dx = ex ∕ {x-1}2 + C
Question 21: e2x sinx
Solution:
Let f(x) = ∫ e2x sinx dx ————– (1)
Integrating by parts, we obtain
f(x) = sinx ∫ e2x dx – ∫ { ❴ d(sinx)∕dx❵ ∫ e2x dx} dx
⇒ f(x) = sinx . e2x∕2 – ∫ cosx e2x∕2 dx
⇒ f(x) = 1∕2 e2x sinx – 1∕2 ∫ e2x cosx dx
Again, Integrating by parts, we obtain
f(x) = 1∕2 e2x sinx – 1∕2 [cosx ∫ e2x dx – ∫ {❴d∕dx cosx❵∫e2x dx} dx]
⇒ f(x) = 1∕2 e2x sinx – 1∕2 [cosx e2x ∕ 2 – ∫ ❴-sinx❵ e2x ∕2 dx
⇒ f(x) = 1∕2 e2x sinx – 1∕2 [(cosx e2x ) ∕ 2 + ∫ sinx e2x ∕2 dx
⇒ f(x) = 1∕2 e2x sinx – (e2x cosx) ∕ 4 – 1∕4f(x) ———-from (1)
⇒ f(x) + 1∕4f(x) = 1∕2 e2x sinx – (e2x cosx) ∕ 4
⇒ 5/4 f(x) = (e2x sinx)1∕2 – (e2x cosx) ∕ 4
⇒ f(x) = e2x ∕5 [ 2 sinx – cosx] + C
Question 22: sin-1(2x∕(1+x2)
Solution:
sin-1(2x∕(1+x2)
Let x = tanθ or dx = sec2θ dθ
So, sin-1(2x∕(1+x2) = sin-1(2 tanθ∕(1+tan2θ) = sin-1(sin2θ) = 2θ
Integrating by parts, we obtain
2[θ.∫sec2θ dθ – ∫{❴dθ∕ dθ❵sec2θ dθ} dθ]
= 2[θ.tanθ – ∫tanθ dθ]
= 2[θ.tanθ – log|cosθ|] + C
= 2[x.tan-1x – log|1√(1+x2|] + C
= 2xtan-1x + 2 log(1+x2)1∕2 + C
= 2x tan-1x + 2 [-1∕2 log(1+x2) ] + C
= 2x tan-1x – log(1+x2) + C
Choose the correct answer in Exercises 23 and 24.
Question 23: ∫ x2ex^3 dx equals
(A) 1∕3 .eX^3 + C
(B) 1∕3 . eX^3 + C
(C) 1∕2 .eX^3 + C
(D) 1∕3 . eX^3 + C
Correct answer is A.
Question 24 :∫ ex secx(1+tanx) dx equals.
(A) ex cosx + C
(B) ex secx + C
(C) ex sinx + C
(D) ex tanx + C
Correct answer is (B) ex secx + C.
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