Evaluate ∫ from 0 to π/2 of 1/(sin x + cos x) dx

• A: √2 ln(√2 + 1)
• B: 1/√2 ln(√2 + 1)
• C: 1
• D: ln(√2 + 1)

To evaluate the integral ∫ from 0 to π/2 of 1/(sin x + cos x) dx, we can use a substitution method. First, notice that sin x + cos x can be rewritten using the identity:

Using a Trigonometric Identity

We know that:

• sin x + cos x = √2 sin(x + π/4)

This allows us to rewrite the integral as:

∫ from 0 to π/2 of 1/(√2 sin(x + π/4)) dx

Changing the Limits

Next, we change the variable:

• Let u = x + π/4, which gives du = dx.
• When x = 0, u = π/4.
• When x = π/2, u = 3π/4.

Transforming the Integral

The integral now becomes:

∫ from π/4 to 3π/4 of 1/(√2 sin u) du

This simplifies to:

(1/√2) ∫ from π/4 to 3π/4 of csc u du

Evaluating the Integral

The integral of csc u is:

-ln |csc u + cot u| + C

Evaluating this from π/4 to 3π/4 gives:

• At u = 3π/4: -ln |csc(3π/4) + cot(3π/4)|
• At u = π/4: -ln |csc(π/4) + cot(π/4)|

Final Calculation

After substituting the values and simplifying, we find:

The result of the integral is:

√2 ln(√2 + 1)

Final Answer

Thus, the value of the integral is:

A: √2 ln(√2 + 1)