Direction: In the following questions, A statement of Assertion (A) is...
∫ex [f(x) + f′(x)]dx = ∫exf(x)dx + ∫exf′(x)dx
= f(x)ex - ∫f′(x)exdx + ∫f′(x)exdx
= ex f(x) + c
Hence R is true.
∫ex(sin x - cos x)dx = ex (-cos) + c
= -ex cos x + c
Hence A is false.
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Direction: In the following questions, A statement of Assertion (A) is...
Assertion (A): ∫ex[sin x cos x] dx = ex sin x
Reason (R): ∫ex [f(x) f′(x)]dx = ex f(x)
The given assertion states that ∫ex[sin x cos x] dx is equal to ex sin x.
The reason states a general rule that ∫ex [f(x) f′(x)]dx is equal to ex f(x).
Explanation:
To evaluate the given integral, let's use integration by parts.
Integration by parts formula is given by:
∫u dv = uv - ∫v du
Let's assume u = ex and dv = sin x cos x dx.
Differentiating u with respect to x, we get:
du/dx = ex
Integrating dv, we get:
v = -1/2 sin^2 x
Using the integration by parts formula, we have:
∫ex[sin x cos x] dx = ex (-1/2 sin^2 x) - ∫(-1/2 sin^2 x)(ex) dx
Simplifying this expression, we get:
∫ex[sin x cos x] dx = -1/2 ex sin^2 x + 1/2 ∫sin^2 x ex dx
Now, let's evaluate the remaining integral.
Using the identity sin^2 x = (1 - cos 2x)/2, we have:
∫sin^2 x ex dx = ∫(1 - cos 2x)/2 ex dx
Splitting the integral, we get:
∫sin^2 x ex dx = 1/2 ∫ ex dx - 1/2 ∫cos 2x ex dx
Integrating the first term, we have:
1/2 ∫ ex dx = 1/2 ex
Integrating the second term, we have:
-1/2 ∫cos 2x ex dx = -1/4 ∫d(ex) cos 2x = -1/4 ex cos 2x + 1/4 ∫ex sin 2x dx
Now, let's evaluate the remaining integral.
Using the identity sin 2x = 2 sin x cos x, we have:
1/4 ∫ex sin 2x dx = 1/4 ∫ex (2 sin x cos x) dx
Splitting the integral, we get:
1/4 ∫ex (2 sin x cos x) dx = 1/2 ∫ex sin x cos x dx
Substituting this result back into the previous expression, we have:
-1/4 ex cos 2x + 1/4 ∫ex sin 2x dx = -1/4 ex cos 2x + 1/2 ∫ex sin x cos x dx
Now, let's substitute this back into the original expression:
∫ex[sin x cos x] dx = -1/2 ex sin^2 x + 1/2 ∫sin^2 x ex dx
= -1/2 ex sin^2 x + 1/2 (-1/4 ex cos 2x + 1/2 ∫ex sin x cos x dx)
Simplifying this expression, we get:
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