Class CBSE Class 12 Mathematics Definite Integrals Q #1306
KNOWLEDGE BASED
REMEMBER
5 Marks 2024 AISSCE(Board Exam) LA
Evaluate: $\int_{0}^{\frac{\pi}{4}}\frac{sin~x+cos~x}{9+16~sin~2x}dx$
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Detailed Solution

Step 1: Rewrite the integrand

We are given the integral $\int_{0}^{\frac{\pi}{4}}\frac{sin~x+cos~x}{9+16~sin~2x}dx$. We can rewrite $sin~2x$ as $1 - (sin~x - cos~x)^2$.

Step 2: Substitute and simplify

Let $t = sin~x - cos~x$. Then, $dt = (cos~x + sin~x)dx$. Also, when $x = 0$, $t = sin~0 - cos~0 = 0 - 1 = -1$. When $x = \frac{\pi}{4}$, $t = sin~\frac{\pi}{4} - cos~\frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$. So, the integral becomes $\int_{-1}^{0}\frac{dt}{9+16(1-t^2)} = \int_{-1}^{0}\frac{dt}{9+16-16t^2} = \int_{-1}^{0}\frac{dt}{25-16t^2}$.

Step 3: Evaluate the integral

We have $\int_{-1}^{0}\frac{dt}{25-16t^2} = \int_{-1}^{0}\frac{dt}{5^2 - (4t)^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}log|\frac{a+x}{a-x}| + C$, we have $\int \frac{dt}{5^2 - (4t)^2} = \frac{1}{4} \int \frac{4dt}{5^2 - (4t)^2} = \frac{1}{4} \cdot \frac{1}{2 \cdot 5} log|\frac{5+4t}{5-4t}|$. So, $\int_{-1}^{0}\frac{dt}{25-16t^2} = \frac{1}{40} [log|\frac{5+4t}{5-4t}|]_{-1}^{0} = \frac{1}{40} [log|\frac{5+0}{5-0}| - log|\frac{5-4}{5+4}|] = \frac{1}{40} [log(1) - log(\frac{1}{9})] = \frac{1}{40} [0 - log(1) + log(9)] = \frac{1}{40} log(9) = \frac{1}{40} log(3^2) = \frac{2}{40} log(3) = \frac{1}{20} log(3)$.

Final Answer: $\frac{1}{20}log(3)$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the standard integral formula and applying a standard substitution technique.
Knowledge Dimension: PROCEDURAL
Justification: The question involves applying a specific procedure of substitution and using a standard integral formula to solve the definite integral.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's ability to apply integration techniques learned in the textbook.

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