Evaluate : _0^(π)/(4) ( x + x)/(9 + 16 2x) dx

MathsCBSE Class 12 · Board exam 2024, set 65/1/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2024: question 33(a)

Set in the CBSE Class 12 board exam of 2024, paper 65/1/1, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerIntegrals

Evaluate : $\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The numerator $sin x + cos x$ is the derivative of $sin x - cos x$ .

Hint 2

Square $sin x - cos x = t$ to write $sin 2 x$ in terms of $t$ .

Hint 3

You should get a standard form $\int \frac{d t}{a ^{2} - t ^{2}}$ with $a = 5/4$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Put $sin x - cos x = t$ , so $(cos x + sin x) d x = d t$ . Limits: $x = 0 \Rightarrow t = - 1$ ; $x = \frac{π}{4} \Rightarrow t = 0$ .
1 mark
Squaring: $sin^{2} x + cos^{2} x - 2 sin x cos x = t^{2} \Rightarrow sin 2 x = 1 - t^{2}$ , so $9 + 16 sin 2 x = 25 - 16 t^{2}$ .
½ mark
$I = \int_{- 1}^{0} \frac{d t}{25 - 16 t ^{2}} = \frac{1}{16} \int_{- 1}^{0} \frac{d t}{( \frac{5}{4} ) ^{2} - t ^{2}}$
1 mark
Using $\int \frac{d t}{a ^{2} - t ^{2}} = \frac{1}{2 a} lo g \frac{a + t}{a - t}$ : $I = \frac{1}{16} \cdot \frac{1}{2 \cdot \frac{5}{4}} [lo g \frac{\frac{5}{4} + t}{\frac{5}{4} - t}]_{- 1}^{0} = \frac{1}{40} [lo g \frac{5 + 4 t}{5 - 4 t}]_{- 1}^{0}$
1½ marks
$= \frac{1}{40} [lo g 1 - lo g \frac{1}{9}] = \frac{1}{40} lo g 9 = \frac{1}{20} lo g 3$
1 mark

Answer: $\frac{1}{40} lo g 9 = \frac{1}{20} lo g 3$

Where the marks are

Substitution $sin x - cos x = t$ with $d t$ and new limits (1 mark)
Expresses $sin 2 x = 1 - t^{2}$ (0.5 mark)
Reduces to $\frac{1}{16} \int_{- 1}^{0} \frac{d t}{( 5/4 ) ^{2} - t ^{2}}$ (1 mark)
Correct standard integration (1.5 marks)
Evaluates to $\frac{1}{20} lo g 3$ (1 mark)

Revise the chapter

Chapter: Integrals (notes, formulas, where marks are lost and every past question)
Topic: Properties of Definite Integrals

Practise this chapterBoard-style questions from Integrals, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2024: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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