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If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ?
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If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 ...
Solution:
Given:
I = ∫(x dx / 1 sin2x cos2x ) from limit 0 to π/4
I = π ln2/b
To find: Value of b
We know that,
I = ∫(x dx / 1 sin2x cos2x ) from limit 0 to π/4
= ∫(x dx / sin2x cos2x ) from limit 0 to π/4
= ∫(x dx / (sinx cosx)^2 ) from limit 0 to π/4
= ∫(x sec2x tan2x dx) from limit 0 to π/4
Now, we can use integration by parts.
Let u = x and dv = sec2x tan2x dx
Then, du/dx = 1 and v = -1/2 tanx
Using the integration by parts formula:
∫u dv = u v - ∫v du
∫(x sec2x tan2x dx) = (-1/2)x tanx + 1/2 ∫tanx dx
= (-1/2)x tanx + 1/2 ln|secx| + C
Now, substituting the limits of integration, we get:
I = π ln2/b
= (-1/2)(π/4) tan(π/4) + 1/2 ln|sec(π/4)| - (-1/2)(0) tan(0) + 1/2 ln|sec(0)|
= (-1/2)(π/4) + 1/2 ln√2
= -π/8 + ln√2
Therefore, we have:
b = e^(-π/8 + ln√2)
= (e^ln√2)/(e^(π/8))
= √2/e^(π/8)
Hence, the value of b is √2/e^(π/8).
Given:
I = ∫(x dx / 1 sin2x cos2x ) from limit 0 to π/4
I = π ln2/b
To find: Value of b
We know that,
I = ∫(x dx / 1 sin2x cos2x ) from limit 0 to π/4
= ∫(x dx / sin2x cos2x ) from limit 0 to π/4
= ∫(x dx / (sinx cosx)^2 ) from limit 0 to π/4
= ∫(x sec2x tan2x dx) from limit 0 to π/4
Now, we can use integration by parts.
Let u = x and dv = sec2x tan2x dx
Then, du/dx = 1 and v = -1/2 tanx
Using the integration by parts formula:
∫u dv = u v - ∫v du
∫(x sec2x tan2x dx) = (-1/2)x tanx + 1/2 ∫tanx dx
= (-1/2)x tanx + 1/2 ln|secx| + C
Now, substituting the limits of integration, we get:
I = π ln2/b
= (-1/2)(π/4) tan(π/4) + 1/2 ln|sec(π/4)| - (-1/2)(0) tan(0) + 1/2 ln|sec(0)|
= (-1/2)(π/4) + 1/2 ln√2
= -π/8 + ln√2
Therefore, we have:
b = e^(-π/8 + ln√2)
= (e^ln√2)/(e^(π/8))
= √2/e^(π/8)
Hence, the value of b is √2/e^(π/8).
Community Answer
If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 ...
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If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ?
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If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ?.
If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If I = integration of (x dx / 1 + sin2x + cos2x ) from limit 0 to π/4 . I = π ln2/b . Then find b ?.
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