Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.11

Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.11

By using properties of definite integrals, evaluate the following:
Question 1.
(int_{0}^{frac{pi}{2}}) cos²x
Solution:

Question 2.
(frac{sqrt{sin x}}{sqrt{sin x}+sqrt{cos x}})dx
Solution:

Question 3.
(int_{0}^{frac{pi}{2}} frac{sin ^{frac{3}{2}} x}{sin ^{frac{3}{2}} x+cos ^{frac{3}{2}} x})dx
Solution:

Question 4.
(int_{0}^{frac{pi}{2}} frac{cos ^{5} x}{sin ^{5} x+cos ^{5} x})
Solution:

Question 5.
(int_{-5}^{5})|x + 2|dx
Solution:

Question 6.
(int_{2}^{8})|x – 5|dx
Solution:

Question 7.
(int_{0}^{1})x(1 – x)ⁿdx
Solution:

Question 8.
(int_{0}^{frac{pi}{2}})log(1 + tan x)dx
Solution:

Question 9.
(int_{0}^{2})x(sqrt{2-x})dx
Solution:

Question 10.
(int_{0}^{frac{pi}{2}})(2log sin x – log sin2x)dx
Solution:

Question 11.
(int_{-frac{pi}{2}}^{frac{pi}{2}} sin ^{2} x d x)
Solution:
Let f(x) = sin²x. Then,
f(- x) = [sin(- x)]² = (- sin x)² = sin²x = f(x).
∴ f(x) is an even function.

Question 12.
(int_{0}^{π}) (frac{xdx}{1+sinx})
Solution:

Adding (1) and (2), we have:

Question 13.
(int_{-frac{pi}{2}}^{frac{pi}{2}} sin ^{7} x d x)
Solution:
Let f(x) = sin⁷x
f(- x) = [sin (- x)]⁷ = (- sin x)⁷ = – sin⁷x = – f(x)
⇒ f(x) is an odd function of x.
But (int_{-a}^{a})f(x) dx = 0 when x is odd.
∴ (int_{frac{pi}{2}}^{frac{pi}{2}} sin ^{7} x) dx = 0.

Question 14.
(int_{0}^{2π}) cos⁵x
Solution:
f(x) = cos⁵x
∴ f(2π – x) = cos5(2π – x) = cos⁵x = f(x).
∴ I = (int_{0}^{2π}) cos⁵x dx = 2(int_{0}^{π})cos⁵x dx.
Again taking g(x) = cos⁵x, we get
g(π – x) = cos⁵(π – x) = – cos⁵x = – g(x)
∴ I = 0. (Because g is an odd function.)
Hence, (int_{0}^{2π})cos⁵x dx = 0.

Question 15.
(int_{0}^{frac{pi}{2}}) (frac{sinx-cosx}{1+sinxcosx})dx
Solution:

Adding (1) and (2), we get

Question 16.
(int_{0}^{π})log(1 + cosx)dx
Solution:

Adding (4) and (5), we get

Put 2x = t so that 2 dx = dt.

Question 17.
(int_{0}^{a} frac{sqrt{x}}{sqrt{x}+sqrt{a-x}})dx
Solution:

Question 18.
(int_{0}^{1})|x – 1|dx
Solution:

Question 19.
Prove that (int_{0}^{a}) f(x)g(x) dx = 2(int_{0}^{a}) f(x)dx,
if f and g are defined as f(x) = f(a – x) and g(x) + g(x – a) = 4.
Solution:

Hence, the result.

Choose the correct answers in questions 20 and 21:
Question 20.
The value of (int_{-frac{pi}{2}}^{frac{pi}{2}})(x³ + x cosx + tan⁵x + 1) dx is
(A) 0
(B) 2
(C) π
(D) 1
Solution:

∴ Part(C) is the correct answer.

Question 21.
The value of (int_{0}^{frac{pi}{2}}) log ((frac{4+3sinx}{4+3cosx})) dx is
(A) 2
(B) (frac{3}{4})
(C) 0
(D) – 2
Solution:

∴ I = 0.
∴ Part(C) is the correct answer.